If the points of contact of the tangents drawn from $(0,0)$ to the curve $y=x^2+3 x+4$ are $(\alpha, \beta)$…
If the points of contact of the tangents drawn from $(0,0)$ to the curve $y=x^2+3 x+4$ are $(\alpha, \beta)$ and $(\gamma, \delta)$, then $\beta+\delta=$
$7$
$25$
$16$
$13$
Solution
Given $y=x^2+3 x+4$
Now $\frac{d y}{d x}=2 x+3$
At $(h, k)$
$\left.\Rightarrow \frac{d y}{d x}\right|_{(h, k)}=2 h+3$
So, equation of tangent at (h, k)
$y-k=(2 h+3)(x-h)$
It is passes through $(0,0)$
So $-k=(2 h+3)(-h) \Rightarrow k=2 h^2+3 k...(i)$
and $k=h^2+3 h+4...(ii)$
After solving (i) \& (ii), we get
$\begin{aligned}
& (h, k)=(-2,2) \text { or }(2,14) \\
& \text { so }(\alpha, \beta)=(-2,2) \\
& (\gamma, \delta)=(2,14) \\
& \Rightarrow \beta+\delta=2+14=16
\end{aligned}$