If the line $x\cos {\rm{\alpha }} + y\sin {\rm{\alpha }} = p$ be normal to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, then

${p^2}\left( {{a^2}{{\sec }^2}{\rm{\alpha }} + {b^2}{\rm{cose}}{{\rm{c}}^2}{\rm{\alpha }}} \right) = {\left( {{a^2} - {b^2}} \right)^2}$

${p^2}\left( {{a^2}{{\cos }^2}{\rm{\alpha }} + {b^2}{{\sin }^2}{\rm{\alpha }}} \right) = {a^2} - {b^2}$

${p^2}\left( {{a^2}{{\cos }^2}{\rm{\alpha }} + {b^2}{{\sin }^2}{\rm{\alpha }}} \right) = {\left( {{a^2} - {b^2}} \right)^2}$

${p^2}\left( {{a^2}{{\sec }^2}{\rm{\alpha }} + {b^2}{\rm{cose}}{{\rm{c}}^2}{\rm{\alpha }}} \right) = {a^2} - {b^2}$

JEE MATHEMATICS MCQ: ELLIPSE

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Analyzing +4 | -1

If the line

x\cos \mathrm{\alpha } + y\sin \mathrm{\alpha } = p

be normal to the ellipse

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1

, then

View answer and explanation

Correct option: D

{p^2}\left( {{a^2}{{\sec }^2}\mathrm{\alpha } + {b^2}\mathrm{cose}{\mathrm{c}^2}\mathrm{\alpha }} \right) = {\left( {{a^2} - {b^2}} \right)^2}

Option D (

{p^2}\left( {{a^2}{{\sec }^2}\alpha + {b^2}\mathrm{cose}{\mathrm{c}^2}\alpha } \right) = {\left( {{a^2} - {b^2}} \right)^2}

) is correct because it is derived by equating the normal equation of the ellipse with the given line equation.