Conics: Co-ordinate Geometry: problem solving for IITJEE Mains: Basics 3

One more “nice” problem with solution ! Though, I strongly suggest you try the problem on your own and compare it with the solution given here. You might even come up with a clever approach than the one I try here:


If r_{1}, r_{2} be the lengths of the radii vectors of the parabola y^{2}=4ax which are drawn at right angles to one another from the vertex. Prove that r_{1}^{4/3}r_{2}^{4/3}=16a^{2}(r_{1}^{2/3}+r_{2}^{2/3}).


Let P(r_{1}\cos{\theta}, r_{2}\sin{\theta}) and Q(r_{2}\sin{\theta},r_{2}\cos{\theta}) be two points on the parabola y^{2}=4ax with lengths of the radii vectors r_{1} and r_{2} respectively, then r_{1}^{2}\sin^{2}{\theta}=4ar_{1}\cos{\theta} \Longrightarrow r_{1}=\frac{4a\cos{\theta}}{\sin^{2}{\theta}}

Similarly, r_{2}=\frac{4a\sin{\theta}}{\cos^{2}{\theta}} since radii vectors r_{1} and r_{2} are at right angles.

Hence, r_{1}r_{2}=\frac{16a^{2}}{\sin{\theta}\cos{\theta}}

\Longrightarrow (r_{1}r_{2})^{4/3}=\frac{(16a^{2})^{4/3}}{(\sin{\theta}\cos{\theta})^{4/3}}…call this equation I.

And, r_{1}^{2/3}+r_{2}^{2/3}=(4a)^{2/3}[\frac{\cos^{2/3}{\theta}}{\sin^{4/3}{\theta}} + \frac{\sin^{2/3}{\theta}}{\cos^{4/3}{\theta}}]

which equals \frac{(4a)^{2/3}}{\sin^{4/3}{\theta}\cos^{4/3}{\theta}}[\cos^{2}{\theta}+\sin^{2}{\theta}]….call this equation II.

From I and II, we get : \frac{r_{1}^{4/3}r_{2}^{4/3}}{r_{1}^{2/3}+r_{2}^{2/3}}=\frac{(16a^{2})^{4/3}}{(16a^{2})^{4/3}}=16a^{2}.

More later,

Nalin Pithwa.

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