IIT-JEE 1995Question 1
Let $d$ be the perpendicular distance from the centre of the ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ to the tangent drawn at a point $P$ on the ellipse. If $F_1$ and $F_2$ are the two foci of the ellipse, then show that $(PF_1-PF_2)^{2}=4a^{2}\left(1-\dfrac{b^{2}}{d^{2}}\right)$.
Ellipse
Answerproved
Solution
Concept. Write both sides in terms of the eccentric angle $t$ of $P$; each becomes $4(a^2-b^2)\cos^2t$. Formulas used. (1) For $P(a\cos t,b\sin t)$: $PF_1=a+ae\cos t$, $PF_2=a-ae\cos t$ (foci $(\mp ae,0)$)(2) Tangent at $P$: $\dfrac{x\cos t}{a}+\dfrac{y\sin t}{b}=1$; distance of $O$ from $lx+my=1$ is $\frac1{\sqrt{l^2+m^2}}$(3) $a^2e^2=a^2-b^2$Step 1 - Why (1) holds. With $b^2=a^2(1-e^2)$: $$PF_2^2=(a\cos t-ae)^2+a^2(1-e^2)\sin^2t=a^2(1-e\cos t)^2$$ (expand: $a^2\cos^2t+a^2\sin^2t-2a^2e\cos t+a^2e^2(1-\sin^2t)$). So $PF_2=a(1-e\cos t)$, and similarly $PF_1=a(1+e\cos t)$. Step 2 - Left side. $PF_1-PF_2=2ae\cos t$, so by (3) $$(PF_1-PF_2)^2=4a^2e^2\cos^2t=4(a^2-b^2)\cos^2t$$ Step 3 - The distance $d$ (2). $$\frac1{d^2}=\frac{\cos^2t}{a^2}+\frac{\sin^2t}{b^2}\ \Rightarrow\ \frac{b^2}{d^2}=\frac{b^2\cos^2t}{a^2}+\sin^2t$$ Step 4 - Right side. $$1-\frac{b^2}{d^2}=(1-\sin^2t)-\frac{b^2\cos^2t}{a^2}=\cos^2t\cdot\frac{a^2-b^2}{a^2}$$ So $4a^2\left(1-\frac{b^2}{d^2}\right)=4(a^2-b^2)\cos^2t$. Step 5 - Conclude. Steps 2 and 4 give the same value, so $(PF_1-PF_2)^2=4a^2\left(1-\dfrac{b^2}{d^2}\right)$. $\blacksquare$ Answer: proved