Question

Assertion :$y^2=4x$ is the equation of a parabola.

Through $(\lambda ,\lambda +1)$, $3$ normals can be drawn to the parabola, if $\lambda <2$

Reason: The point $(\lambda ,\lambda +1)$ lies outside the parabola for all $\lambda \ne 1$.

• 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. Assertion is incorrect but reason is correct

Answer 1

The correct option is D Assertion is incorrect but reason is correct

Equation of a normal to the parabola is $y+tx=2t+t^3$,
If it passes through $(\lambda ,\lambda +1)$
then $\lambda +1+t\lambda =2t+t^3$
or $t^3+(2-\lambda )t-(\lambda +1)=0=f\left(t\right)$ say.
Now $f^{\prime }\left(t\right)=3t^2+(2-\lambda )>0$ as $\lambda <2$.
$\therefore f\left(t\right)=0$ has only one real root and thus there is only one normal through $(\lambda ,\lambda +1)$ to the parabola can be drawn.

Thus, statement-1 is false.
Statement-2 is correct as $S\equiv y^2-4x={(\lambda +1)}^2-4\lambda =(\lambda -1{)}^2>0$ if $\lambda \ne 1$
which shows that $(\lambda ,\lambda +1)$ lies outside the parabola $y^2=4x$, if $\lambda \ne -1$.