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

We have y = -x cot $\alpha$ + p cosec $\alpha$ ...(i) Now, line (i) is normal to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}}=1$ , if $c^{2}=\frac{\left(a^{2}-b^{2}\right)^{2} m^{2}}{a^{2}+b^{2} m^{2}}$ $\therefore p^{2} \operatorname{cosec}^{2} \alpha=\frac{\left(a^{2}-b^{2}\right)^{2} \cot ^{2} \alpha}{a^{2}+b^{2} \cot ^{2} \alpha}$ After solving, we get p 2 (a 2 sec 2 $\alpha$ + b 2 cosec 2 $\alpha$ ) = (a 2 - b 2 ) 2

Ellipse

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Question:

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

p2(a2 sec2 $\alpha$ + b2 cosec2 $\alpha$) = (a2 - b2)2

p2(a2 cos2 $\alpha$ + b2 sin2 $\alpha$) = a2 - b2

Step-by-Step Solution

Key Concept: Convert the given line into slope-intercept form to apply the standard condition for normality to an ellipse.

We have y = -x cot $\alpha$ + p cosec$\alpha$ ...(i) Now, line (i) is normal to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}}=1$, if $c^{2}=\frac{\left(a^{2}-b^{2}\right)^{2} m^{2}}{a^{2}+b^{2} m^{2}}$ $\therefore p^{2} \operatorname{cosec}^{2} \alpha=\frac{\left(a^{2}-b^{2}\right)^{2} \cot ^{2} \alpha}{a^{2}+b^{2} \cot ^{2} \alpha}$ After solving, we get p2(a2 sec2 $\alpha$ + b2 cosec2 $\alpha$) = (a2 - b2)2

Correct Answer: A

Question:

Step-by-Step Solution

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