Question

Let $f^{\prime \prime }\left(x\right)+f^{\prime }\left(x\right)+\left(f\right(x){)}^2=x^2$ be the differential equation of a curve $y=f\left(x\right)$ and let $P$ be the point of local maximum of $y=f\left(x\right).$ Then the number of tangents which can be drawn from $P$ to $x^2-y^2=16$ is

Answer 1

At point $P\equiv (x_1,y_1)$ of local maximum,
$f^{\prime }\left(x_1\right)=0$ and $f^{\prime \prime }\left(x_1\right)<0$
$\Rightarrow f^{\prime \prime }\left(x_1\right)=x_1^2-\left(f\right(x_1){)}^2=x_1^2-y_1^2<0$
Now ,
$x_1^2-y_1^2-16<-16<0$
$\Rightarrow$ Point $P$ lies outside the hyperbola.
Hence, $2$ tangents can be drawn.