If the points of contact of the tangents draw | vedclass

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(C) Let the point of contact be $(h, k)$. Since $(h, k)$ lies on the curve $y=x^2+3x+4$,we have $k=h^2+3h+4$ ... $(i)$.The slope of the tangent at $(h

Vedclass · Accepted Answer

$16$

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(C) Let the point of contact be $(h, k)$. Since $(h, k)$ lies on the curve $y=x^2+3x+4$,we have $k=h^2+3h+4$ ... $(i)$.The slope of the tangent at $(h, k)$ is $\frac{dy}{dx} = 2x+3$. At $(h, k)$,the slope is $m = 2h+3$.The equation of the tangent at $(h, k)$ is $y-k = (2h+3)(x-h)$.Since the tangent passes through $(0, 0)$,we substitute $x=0$ and $y=0$:$0-k = (2h+3)(0-h) \Rightarrow -k = -2h^2-3h \Rightarrow k = 2h^2+3h$ ... $(ii)$.Equating $(i)$ and $(ii)$:$h^2+3h+4 = 2h^2+3h$$h^2 = 4 \Rightarrow h = \pm 2$.If $h = -2$,$k = (-2)^2+3(-2)+4 = 4-6+4 = 2$. So,$(\alpha, \beta) = (-2, 2)$.If $h = 2$,$k = (2)^2+3(2)+4 = 4+6+4 = 14$. So,$(\gamma, \delta) = (2, 14)$.Thus,$\beta+\delta = 2+14 = 16$.

If the points of contact of the tangents drawn from $(0,0)$ to the curve $y=x^2+3x+4$ are $(\alpha, \beta)$ and $(\gamma, \delta)$,then $\beta+\delta=$

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