Question

Assertion :A line $2px+y\sqrt{1-p^2}=1$ $\forall p\in [-1,1]$ touches a fixed ellipse then eccentricity of the ellipse is $\frac{\sqrt{3}}{2}$ Reason: The eccentricity of an ellipse is evaluated by the formula $b^2=a^2(1-e^2)$ if $b<a$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Tangent of an ellipse with centre at $(0,0)$ is $\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$

Compare with $2px+y\sqrt{1-p^2}=1$

We get $2=\frac{1}{a},b=1$ and let $p$ is $\cos \theta$

$\frac{a}{b}=\frac{1}{2}$

So the ellipse is $4x^2+y^2=1$

$a^2=b^2(1-e^2)$

For ellipse

$a^2=b^2(1-e^2)$ if $b>a$

And $b^2=a^2(1-e^2)$ if $b<a$

As $b>a$

So $a^2=b^2(1-e^2)$

$\frac{b^2}{4}=b^2(1-e^2)$

$\frac{1}{4}=1-e^2$

$\frac{3}{4}=e^2$

So $\frac{\sqrt{3}}{2}=e$

So assertion and reason are correct and reason is the correct explanation of assertion.