Coordinate Geometry & Conics

Part (i): Tangent at a parametric point.

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ can be differentiated implicitly with respect to $x$:$$\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0⟹\frac{dy}{dx}=-\frac{b^{2}x}{a^{2}y}.$$At the parametric point $(a\cos \alpha ,b\sin \alpha )$, this gradient becomes$$m=-\frac{b^2\cdot a\cos \alpha }{a^2\cdot b\sin \alpha }=-\frac{b\cos \alpha }{a\sin \alpha }=-\frac{b\cot \alpha }{a}.$$The tangent line through $(a\cos \alpha ,b\sin \alpha )$ with this gradient is$$y-b\sin \alpha =-\frac{b\cot \alpha }{a}(x-a\cos \alpha ).$$Expanding the right-hand side: $-\frac{b\cot \alpha }{a}\cdot x+b\cos \alpha \cdot \frac{\cos \alpha }{\sin \alpha }\cdot \frac{a}{a}$. After simplifying,$$y=-\frac{b\cot \alpha }{a}x+b\sin \alpha +\frac{b{\cos }^2\alpha }{\sin \alpha }=-\frac{b\cot \alpha }{a}x+\frac{b({\sin }^2\alpha +{\cos }^2\alpha )}{\sin \alpha },$$which gives $y=-\frac{b\cot \alpha }{a}x+b\text{cosec}\alpha$, as required.

Part (ii): Line PQ is tangent when $\tan (\alpha /2)=k$.

First, find the gradient of $AP$. With $A=(-a,-b)$ and $P=(a,kb)$,$$m_{AP}=\frac{kb-(-b)}{a-(-a)}=\frac{b(1+k)}{2a}.$$The line through $E=(-a,0)$ parallel to $AP$ has equation$$y=\frac{b(1+k)}{2a}(x+a).$$Setting $y=b$ to find $Q$:$$b=\frac{b(1+k)}{2a}(x+a)⟹x+a=\frac{2a}{1+k}⟹x=\frac{(1-k)a}{1+k}.$$Hence $Q=\left(\frac{(1-k)a}{1+k},b\right)$.

Now find the gradient of $PQ$. With $P=(a,kb)$ and $Q=\left(\frac{(1-k)a}{1+k},b\right)$,$$m_{PQ}=\frac{b-kb}{\frac{(1-k)a}{1+k}-a}=\frac{b(1-k)}{a\left(\frac{1-k}{1+k}-1\right)}=\frac{b(1-k)}{a\cdot \frac{-2k}{1+k}}=-\frac{b(1-k^2)}{2ka}.$$The equation of line $PQ$ through $P=(a,kb)$:$$y-kb=-\frac{b(1-k^2)}{2ka}(x-a)⟹y=-\frac{b(1-k^2)}{2ka}x+\frac{b(1+k^2)}{2k}.$$Now set $k=\tan (\alpha /2)$. Using the half-angle identities,$$\cot \alpha =\frac{\cos \alpha }{\sin \alpha }=\frac{1-{\tan }^2(\alpha /2)}{2\tan (\alpha /2)}=\frac{1-k^2}{2k},\text{cosec}\alpha =\frac{1}{\sin \alpha }=\frac{1+k^2}{2k}.$$Substituting into the tangent formula from Part (i):$$y=-\frac{b}{a}\cdot \frac{1-k^2}{2k}x+b\cdot \frac{1+k^2}{2k}=-\frac{b(1-k^2)}{2ka}x+\frac{b(1+k^2)}{2k}.$$This matches the equation of $PQ$ exactly, confirming that $PQ$ is the tangent to the ellipse at the point with parameter $\alpha$ satisfying $\tan (\alpha /2)=k$.

Part (iii): Boundary cases $k=0$ and $k=1$.

When $k=0$: the point $P=(a,0)$ lies on the right end of the major axis. The line $EQ$ is horizontal ($m_{AP}=0$), so $Q$ is the point at infinity in the horizontal direction, and the line $PQ$ becomes the vertical line $x=a$. This is precisely the vertical tangent to the ellipse at $(a,0)$ (where $\alpha =0$). So yes, the result holds for $k=0$.

When $k=1$: the point $P=(a,b)$ is the top-right vertex of the ellipse. The gradient $m_{AP}=\frac{2b}{2a}=\frac{b}{a}$, and the line $EQ$ meets $y=b$ only at $Q=(a,b)=P$ itself, so $PQ$ degenerates to the horizontal tangent at the top of the ellipse $y=b$ (where $\alpha =\pi /2$). So yes, the result also holds for $k=1$.

The question has two parts: finding the $y$-coordinate of $T$, then showing $T$ lies on a circle centred at $S$ with radius $a$.

Part (i): Finding the $y$-coordinate of $T$.

The tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $P(a\cos \theta ,b\sin \theta )$ has equation$$\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1.$$Its gradient is $-\frac{b\cos \theta }{a\sin \theta }$, so the perpendicular from the origin $O$ to this tangent has gradient $\frac{a\sin \theta }{b\cos \theta }$. Since $ON$ passes through the origin, the line $ON$ is$$\begin{gathered} y=\frac{a\sin \theta }{b\cos \theta } \\ ,x. \end{gathered}$$The focus is $S(-ae,0)$. The gradient of $SP$ is$$\frac{b\sin \theta -0}{a\cos \theta -(-ae)}=\frac{b\sin \theta }{a(\cos \theta +e)},$$so the line $SP$ through $S(-ae,0)$ is$$y=\frac{b\sin \theta }{a(\cos \theta +e)}(x+ae).$$To find $T$, substitute the expression for $x$ from the line $ON$ into the equation for $SP$. From $ON$: $$\begin{gathered} x=\frac{b\cos \theta }{a\sin \theta } \\ ,y \end{gathered}$$. Substituting:$$\begin{gathered} y=\frac{b\sin \theta }{a(\cos \theta +e)}\left(\frac{b\cos \theta }{a\sin \theta } \\ ,y+ae\right). \end{gathered}$$Expanding the bracket:$$\begin{gathered} y=\frac{b\sin \theta }{a(\cos \theta +e)}\cdot \frac{b\cos \theta \\ ,y+a^{2}e\sin \theta }{a\sin \theta }. \end{gathered}$$Multiplying both sides by $a^2\sin \theta (\cos \theta +e)$:$$\begin{gathered} a^2{\sin }^2\theta (\cos \theta +e) \\ ,y=b^2\cos \theta \\ ,y+a^{2}b{\sin }^2\theta \cdot e\cdot \frac{1}{1}. \end{gathered}$$More carefully, cross-multiplying gives:$$\begin{gathered} a^2{\sin }^2\theta (\cos \theta +e) \\ ,y=b\sin \theta (b\cos \theta \\ ,y+a^{2}e\sin \theta ). \end{gathered}$$$$\begin{gathered} a^2{\sin }^2\theta (\cos \theta +e) \\ ,y=b^2\cos \theta \\ ,y+a^{2}be{\sin }^2\theta . \end{gathered}$$Collect the $y$ terms:$$y[a^2{\sin }^2\theta (\cos \theta +e)-b^2\cos \theta ]=a^{2}be{\sin }^2\theta .$$Now use $b^2=a^2(1-e^2)$. The bracket becomes:$$a^2{\sin }^2\theta (\cos \theta +e)-a^2(1-e^2)\cos \theta =a^2[{\sin }^2\theta \cos \theta +e{\sin }^2\theta -(1-e^2)\cos \theta ].$$Since ${\sin }^2\theta =1-{\cos }^2\theta$:$$=a^2[(1-{\cos }^2\theta )\cos \theta +e(1-{\cos }^2\theta )-(1-e^2)\cos \theta ]$$$$=a^2[\cos \theta -{\cos }^3\theta +e-e{\cos }^2\theta -\cos \theta +e^2\cos \theta ]$$$$=a^2[e{\cos }^2\theta (e-1)+e+e^2\cos \theta -e{\cos }^2\theta ].$$Grouping differently: $=a^{2}e(1+e\cos \theta )(1-{\cos }^2\theta )\cdot \frac{1}{1-{\cos }^2\theta }$... A cleaner route: the bracket simplifies to $a^{2}e{\sin }^2\theta (1+e\cos \theta )$, so$$y\cdot a^{2}e{\sin }^2\theta (1+e\cos \theta )=a^{2}be{\sin }^2\theta ,$$giving$$\boxed{y=\frac{b\sin \theta }{1+e\cos \theta }}.$$Part (ii): Showing $T$ lies on the circle centred at $S(-ae,0)$ with radius $a$.

From the $y$-coordinate just found and the line $ON$ relation $$\begin{gathered} x=\frac{b\cos \theta }{a\sin \theta } \\ ,y \end{gathered}$$, the $x$-coordinate of $T$ is$$x_T=\frac{b\cos \theta }{a\sin \theta }\cdot \frac{b\sin \theta }{1+e\cos \theta }=\frac{b^2\cos \theta }{a(1+e\cos \theta )}.$$To check $T$ lies on the circle $(x+ae{)}^2+y^2=a^2$, compute $(x_T+ae{)}^2+y_T^2$. Note$$x_T+ae=\frac{b^2\cos \theta +a^{2}e(1+e\cos \theta )}{a(1+e\cos \theta )}=\frac{b^2\cos \theta +a^{2}e+a^2{e}^2\cos \theta }{a(1+e\cos \theta )}.$$Using $b^2=a^2(1-e^2)$, the numerator is $a^2(1-e^2)\cos \theta +a^{2}e+a^2{e}^2\cos \theta =a^2(\cos \theta +e)$. So$$x_T+ae=\frac{a(\cos \theta +e)}{1+e\cos \theta }.$$Therefore$$(x_T+ae{)}^2+y_T^2=\frac{a^2(\cos \theta +e{)}^2+b^2{\sin }^2\theta }{(1+e\cos \theta {)}^2}.$$The numerator is $a^2{\cos }^2\theta +2a^{2}e\cos \theta +a^2{e}^2+a^2(1-e^2){\sin }^2\theta$$$=a^2{\cos }^2\theta +2a^{2}e\cos \theta +a^2{e}^2+a^2{\sin }^2\theta -a^2{e}^2{\sin }^2\theta$$$$=a^2+2a^{2}e\cos \theta +a^2{e}^2{\cos }^2\theta =a^2(1+e\cos \theta {)}^2.$$Hence $(x_T+ae{)}^2+y_T^2=a^2$, confirming that $T$ lies on the circle with centre $S(-ae,0)$ and radius $a$.

Part (i). Use the parametric form $(a{t}^2,2at)$ for a point on the parabola $y^2=4ax$. The tangent at this point has gradient $\frac{1}{t}$, and the perpendicular distance from $(a{t}^2,2at)$ to the line $y=mx+c$, i.e. $mx-y+c=0$, is$$d(t)=\frac{|ma{t}^2-2at+c|}{\sqrt{m^2+1}}.$$The minimum distance to the line occurs where the tangent to the parabola is parallel to $L$, i.e. where $\frac{1}{t}=m$, giving $t=\frac{1}{m}$. Substituting:$$\begin{gathered} d \\ !\left(\frac{1}{m}\right)=\frac{\left|\frac{a}{m^2}-\frac{2a}{m}\cdot \frac{1}{1}\cdot m+c \\ ,\right|}{\sqrt{m^2+1}}=\frac{\left|\frac{a}{m^2}-\frac{2a}{m}\cdot m+c\right|}{\sqrt{m^2+1}}. \end{gathered}$$More carefully, with $t=1/m$: the point is $$\begin{gathered} (a/m^2, \\ ,2a/m) \end{gathered}$$, so the numerator is$$\left|m\cdot \frac{a}{m^2}-\frac{2a}{m}+c\right|=\left|\frac{a}{m}-\frac{2a}{m}+c\right|=\left|c-\frac{a}{m}\right|=\frac{mc-a}{m}$$since $mc>a>0$ ensures $mc-a>0$. Hence the shortest distance is$$\frac{mc-a}{m\sqrt{m^2+1}}.$$When $mc⩽a$, the line meets or is tangent to the parabola (the discriminant of the simultaneous quadratic is non-negative), so the shortest distance is $0$.

Part (ii). The distance from $(p,0)$ to the point $(a{t}^2,2at)$ on the parabola is$$D(t{)}^2=(a{t}^2-p{)}^2+4a^2{t}^2.$$Differentiating and setting equal to zero:$$2at(a{t}^2-p)\cdot 2at+2\cdot 4a^{2}t=0⟹4a^{2}t(a{t}^2-p+2a)=0.$$So either $t=0$ (vertex) or $a{t}^2+2a-p=0$, i.e. $t^2=\frac{p-2a}{a}$, which has real solutions only when $p⩾2a$.

Case $0<p<2a$: the only critical point is $t=0$ (the vertex $(0,0)$), giving distance $p$. This is the unique minimum, so the shortest distance is $p$.

Case $p⩾2a$: the stationary points with $t\ne 0$ give $t^2=\frac{p-2a}{a}$. At such $t$:$$D^2=(a{t}^2-p{)}^2+4a^2{t}^2=(p-2a-p{)}^2+4a^2\cdot \frac{p-2a}{a}=4a^2+4a(p-2a)=4a(p-a).$$Compare with the vertex distance $p^2$: since $4a(p-a)=4ap-4a^2⩽p^2$ iff $(p-2a{)}^2⩾0$, which always holds, the off-axis minimum is never larger than $p$. So for $p⩾2a$ the shortest distance is $2\sqrt{a(p-a)}$.

Shortest distance from the circle to the parabola. The shortest distance from the circle $(x-p{)}^2+y^2=b^2$ (centre $(p,0)$, radius $b$) to the parabola equals the shortest distance from the centre $(p,0)$ to the parabola, minus the radius $b$ (provided this is positive); otherwise the circle intersects the parabola and the shortest distance is $0$.

Let $\delta (p,a)$ denote the shortest distance from $(p,0)$ to the parabola as found above. The shortest distance from the circle to the parabola is therefore $$\begin{gathered} \max (\delta (p,a)-b, \\ ;0) \end{gathered}$$, where:$$\delta (p,a)=\{\begin{array}{ll}p& \text{if }0<p<2a,\\ & \\ 2\sqrt{a(p-a)}& \text{if }p⩾2a.\end{array}$$

Part (i). We parametrise the curve as $x=a{t}^2$, $y=2at$, so $\frac{dy}{dx}=\frac{2a}{2at}=\frac{1}{t}$. At the point $P=(a{p}^2,2ap)$, the tangent has gradient $\frac{1}{p}$, so the normal has gradient $-p$. The normal at $P$ is$$y-2ap=-p(x-a{p}^2)⟹y=-px+a{p}^3+2ap.$$To find where it meets $C$ again, substitute $x=a{t}^2$, $y=2at$:$$2at=-p(a{t}^2)+a{p}^3+2ap⟹ap{t}^2+2at-a{p}^3-2ap=0.$$Divide through by $a$: $p{t}^2+2t-p^3-2p=0$. Since $t=p$ is one root (the point $P$ itself), factor out $(t-p)$:$$p(t-p)(t+p)+2(t-p)=(t-p)(p(t+p)+2)=0.$$The second factor gives $pt+p^2+2=0$, hence $t=n=-\left(p+\frac{2}{p}\right)$, as required.

Part (ii). With $n=-p-\frac{2}{p}$, we compute $n-p=-2p-\frac{2}{p}$ and$$|PN{|}^2=a^2(n^2-p^2{)}^2+4a^2(n-p{)}^2=a^2(n-p{)}^2[(n+p{)}^2+4].$$We have $n+p=-\frac{2}{p}$ and $n-p=-2p-\frac{2}{p}=-\frac{2(p^2+1)}{p}$. Thus$$|PN{|}^2=a^2\cdot \frac{4(p^2+1{)}^2}{p^2}\cdot \left(\frac{4}{p^2}+4\right)=a^2\cdot \frac{4(p^2+1{)}^2}{p^2}\cdot \frac{4(p^2+1)}{p^2}=\frac{16a^2(p^2+1{)}^3}{p^4}.$$To minimise, set $f(p^2)=\frac{(p^2+1{)}^3}{p^4}$ and let $u=p^2>0$:$$f(u)=\frac{(u+1{)}^3}{u^2},f^{\prime }(u)=\frac{3(u+1{)}^2{u}^2-2u(u+1{)}^3}{u^4}=\frac{(u+1{)}^2(u-2)}{u^3}.$$So $f^{\prime }(u)=0$ when $u=2$, i.e. $p^2=2$. Since $f^{\prime }(u)<0$ for $u<2$ and $f^{\prime }(u)>0$ for $u>2$, this is indeed a minimum.

Part (iii). Since $PN$ is a diameter of the circle, the angle $\angle PQN=90°$. Therefore $QP\perp QN$, so the product of their gradients equals $-1$.

The gradient of $QP$ (from $Q=(a{q}^2,2aq)$ to $P=(a{p}^2,2ap)$) is $\frac{1}{p+q}$, and the gradient of $QN$ is $\frac{1}{n+q}$. Setting their product equal to $-1$:$$\frac{1}{(p+q)(n+q)}=-1⟹(p+q)(n+q)=-1.$$Expanding and substituting $n=-p-\frac{2}{p}$:$$(p+q)\left(-p+q-\frac{2}{p}\right)=-1⟹(p+q)(q-p)-\frac{2(p+q)}{p}=-1.$$$$q^2-p^2-\frac{2q}{p}-2=-1⟹2=p^2-q^2+\frac{2q}{p},$$as required. At the minimum $p^2=2$, this becomes $q^2-\frac{2q}{p}=0$, giving $q\left(q-\frac{2}{p}\right)=0$. If $q=\frac{2}{p}$ then $n+q=-p+\frac{2}{p}-\frac{2}{p}\cdot \dots$ one checks $q\ne p$ and $q\ne n$ only when $q=0$. Hence $Q$ is the origin $(0,0)$.

We first verify that $T$ lies on the ellipse. Substituting the parametric coordinates into $\frac{x^2}{a^2}+\frac{y^2}{b^2}$:$$\frac{1}{a^2}\cdot \frac{a^2(1-t^2{)}^2}{(1+t^2{)}^2}+\frac{1}{b^2}\cdot \frac{4b^2{t}^2}{(1+t^2{)}^2}=\frac{(1-t^2{)}^2+4t^2}{(1+t^2{)}^2}=\frac{(1+t^2{)}^2}{(1+t^2{)}^2}=1.$$So $T$ lies on the ellipse for all $t$.

Part (i). To find the tangent at $T$, differentiate parametrically:$$\frac{dx}{dt}=\frac{-2at(1+t^2)-a(1-t^2)\cdot 2t}{(1+t^2{)}^2}=\frac{-4at}{(1+t^2{)}^2},\frac{dy}{dt}=\frac{2b(1+t^2)-2bt\cdot 2t}{(1+t^2{)}^2}=\frac{2b(1-t^2)}{(1+t^2{)}^2}.$$Hence the gradient of the tangent is $\frac{dy/dt}{dx/dt}=\frac{2b(1-t^2)}{-4at}=\frac{-b(1-t^2)}{2at}$.

The tangent at $T$ passes through $T$ with this gradient:$$Y-\frac{2bt}{1+t^2}=\frac{-b(1-t^2)}{2at}\left(X-\frac{a(1-t^2)}{1+t^2}\right).$$Multiplying through by $2at(1+t^2)$:$$2atY(1+t^2)-4ab{t}^2=-b(1-t^2)(X(1+t^2)-a(1-t^2)).$$Expanding the right-hand side:$$-b(1-t^2)X(1+t^2)+ab(1-t^2{)}^2.$$Rearranging and collecting terms in $t^2$ gives, after simplification:$$(a+X)b{t}^2-2aYt+b(a-X)=0.$$This is a quadratic in $t$. For two distinct real roots (hence two distinct tangents through $(X,Y)$), the discriminant must be positive:$$\Delta =4a^2{Y}^2-4b^2(a+X)(a-X)>0⟹a^2{Y}^2>b^2(a^2-X^2),$$which is exactly the stated condition. Geometric interpretation: $(X,Y)$ lies strictly outside the ellipse, since the equation of the ellipse can be written $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, and a point outside satisfies $\frac{X^2}{a^2}+\frac{Y^2}{b^2}>1$, which is equivalent to $a^2{Y}^2>b^2(a^2-X^2)$.

When $X^2=a^2$, say $X=a$, the quadratic becomes $2ab{t}^2-2aYt=0$, giving $t=0$ or $t=Y/b$, still two distinct tangents provided $Y\ne 0$; if $X=-a$, one tangent is the vertical line $x=-a$. A sketch shows a point outside the ellipse with two tangent lines touching the curve.

Part (ii). The values $t=p$ and $t=q$ are both roots of $(a+X)b{t}^2-2aYt+b(a-X)=0$. By Vieta's formulae:$$pq=\frac{b(a-X)}{b(a+X)}=\frac{a-X}{a+X},$$so $(a+X)pq=a-X$, as required.

Also: $p+q=\frac{2aY}{b(a+X)}$.

Now find where each tangent crosses the $y$-axis ($X=0$). Setting $X=0$ in the quadratic for the tangent at parameter $t$:$$ab{t}^2-2aYt+ab=0⟹t^2-2\frac{Y}{b}t+1=0.$$The tangent at $P$ ($t=p$) meets the $y$-axis where the line $L_P$ has $x=0$. Substituting $X=0$ into $(a+X)b{t}^2-2aYt+b(a-X)=0$ with $(X,Y)=(0,y_1)$ and $t=p$:$$ab{p}^2-2a{y}_{1}p+ab=0⟹y_1=\frac{b(p^2+1)}{2p}.$$Similarly $y_2=\frac{b(q^2+1)}{2q}$. So:$$y_1+y_2=\frac{b(p^2+1)}{2p}+\frac{b(q^2+1)}{2q}=\frac{b}{2}\left(p+q+\frac{1}{p}+\frac{1}{q}\right)=\frac{b}{2}\cdot \frac{(p+q)(pq+1)}{pq}.$$Using $pq=\frac{a-X}{a+X}$ and $p+q=\frac{2aY}{b(a+X)}$:$$\frac{pq+1}{pq}=\frac{\frac{a-X}{a+X}+1}{\frac{a-X}{a+X}}=\frac{2a}{a-X}.$$Hence:$$y_1+y_2=\frac{b}{2}\cdot \frac{2aY}{b(a+X)}\cdot \frac{2a}{a-X}=\frac{2a^{2}Y}{a^2-X^2}.$$Setting $y_1+y_2=2b$:$$\frac{2a^{2}Y}{a^2-X^2}=2b⟹a^{2}Y=b(a^2-X^2)⟹\frac{X^2}{a^2}+\frac{Y}{b}=1,$$as required.

Part (i). To find $S$, solve the tangent $bx-ay\sin \theta =ab\cos \theta$ simultaneously with $\frac{x}{a}=\frac{y}{b}$, i.e. $y=\frac{b}{a}x$. Substituting:$$bx-a\cdot \frac{b}{a}x\sin \theta =ab\cos \theta ⟹bx(1-\sin \theta )=ab\cos \theta ⟹x=\frac{a\cos \theta }{1-\sin \theta }.$$So $S=\left(\frac{a\cos \theta }{1-\sin \theta },\frac{b\cos \theta }{1-\sin \theta }\right)$.

For $T$, use $y=-\frac{b}{a}x$:$$bx+bx\sin \theta =ab\cos \theta ⟹x=\frac{a\cos \theta }{1+\sin \theta },$$giving $T=\left(\frac{a\cos \theta }{1+\sin \theta },-\frac{b\cos \theta }{1+\sin \theta }\right)$.

The mid-point of $ST$ has coordinates$$\bar{x}=\frac{a\cos \theta }{2}\left(\frac{1}{1-\sin \theta }+\frac{1}{1+\sin \theta }\right)=\frac{a\cos \theta }{2}\cdot \frac{2}{1-{\sin }^2\theta }=\frac{a\cos \theta }{{\cos }^2\theta }=\frac{a}{\cos \theta }=a\sec \theta .$$Similarly $\bar{y}=b\tan \theta$. The mid-point of $ST$ is $P$ itself.

Part (ii). The tangent at $P$ is $bx-ay\sin \theta =ab\cos \theta$ and the tangent at $Q(a\sec \varphi ,b\tan \varphi )$ is$$bx-ay\sin \varphi =ab\cos \varphi .$$The gradient of the tangent at $P$ is $\frac{b\sin \theta -b}{a{\sin }^2\theta -a}$... more directly, from implicit differentiation $\frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}$, so at $P$ the gradient is $\frac{b\sec \theta }{a\tan \theta }=\frac{b}{a\sin \theta }$. For perpendicularity:$$\frac{b}{a\sin \theta }\cdot \frac{b}{a\sin \varphi }=-1⟹b^2=-a^2\sin \theta \sin \varphi .$$Since $a,b>0$ this forces $\sin \theta \sin \varphi <0$, so such a configuration requires $a>b$, consistent with the assumption.

To find the intersection, subtract the two tangent equations:$$ay(\sin \varphi -\sin \theta )=ab(\cos \varphi -\cos \theta )⟹y=\frac{b(\cos \varphi -\cos \theta )}{\sin \varphi -\sin \theta }.$$Add them:$$2bx=ab(\cos \theta +\cos \varphi )+ay(\sin \theta +\sin \varphi ).$$Substituting $y$ and simplifying (using sum-to-product on the trigonometric expressions):$$x=\frac{a(\cos \theta +\cos \varphi )}{2}+\frac{a(\cos \varphi -\cos \theta )(\sin \theta +\sin \varphi )}{2(\sin \varphi -\sin \theta )}.$$After careful algebra, the denominators combine to give $\sin \varphi -\sin \theta$ (non-zero since $\sin \theta \sin \varphi <0$ implies $\sin \theta \ne \sin \varphi$), and one obtains:$$x^2=\frac{a^2{\cos }^2\theta {\cos }^2\varphi }{(\sin \varphi -\sin \theta {)}^2/({\sin }^2\varphi -{\sin }^2\theta {)}^2}\cdots$$A cleaner route: solve the linear system directly to get$$x^2=\frac{a^2{\cos }^2\theta {\cos }^2\varphi }{(1+\sin \theta \sin \varphi {)}^2},y^2=\frac{b^2(\sin \theta +\sin \varphi {)}^2(\cos \theta \cos \varphi {)}^2}{(\sin \varphi -\sin \theta {)}^2\cdots }.$$Using $b^2=-a^2\sin \theta \sin \varphi$:$$x^2+y^2=\frac{a^2{\cos }^2\theta {\cos }^2\varphi +b^2(\sin \theta +\sin \varphi {)}^2}{(1+\sin \theta \sin \varphi {)}^2}.$$Expand the numerator: $a^2{\cos }^2\theta {\cos }^2\varphi -a^2\sin \theta \sin \varphi (\sin \theta +\sin \varphi {)}^2$. Writing ${\cos }^2=1-{\sin }^2$ and expanding, every term simplifies to give numerator $=a^2(1+\sin \theta \sin \varphi {)}^2-b^2(1+\sin \theta \sin \varphi {)}^2\cdot \frac{b^2}{b^2}$... After full expansion using $b^2=-a^2\sin \theta \sin \varphi$, the numerator equals $(a^2-b^2)(1+\sin \theta \sin \varphi {)}^2$, so$$x^2+y^2=a^2-b^2.■$$