The equation of common tangents to the parabo | vedclass

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Question

(A) The equation of a tangent to the parabola $y^2 = 8x$ (where $4a = 8$,so $a = 2$) is given by $y = mx + \frac{2}{m}$.The equation of the hyperbola

Vedclass · Accepted Answer

$2x \pm y + 1 = 0$

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(A) The equation of a tangent to the parabola $y^2 = 8x$ (where $4a = 8$,so $a = 2$) is given by $y = mx + \frac{2}{m}$.The equation of the hyperbola is $3x^2 - y^2 = 3$,which can be written as $\frac{x^2}{1} - \frac{y^2}{3} = 1$. Here $a^2 = 1$ and $b^2 = 3$.The condition for the line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.Substituting $c = \frac{2}{m}$,$a^2 = 1$,and $b^2 = 3$ into the condition:$(\frac{2}{m})^2 = 1(m^2) - 3$$\frac{4}{m^2} = m^2 - 3$$4 = m^4 - 3m^2$$m^4 - 3m^2 - 4 = 0$$(m^2 - 4)(m^2 + 1) = 0$Since $m$ must be real,$m^2 = 4$,so $m = \pm 2$.Substituting $m = \pm 2$ into the tangent equation $y = mx + \frac{2}{m}$:For $m = 2$,$y = 2x + \frac{2}{2} \implies y = 2x + 1 \implies 2x - y + 1 = 0$.For $m = -2$,$y = -2x + \frac{2}{-2} \implies y = -2x - 1 \implies 2x + y + 1 = 0$.Thus,the common tangents are $2x \pm y + 1 = 0$.

The equation of common tangents to the parabola $y^2 = 8x$ and the hyperbola $3x^2 - y^2 = 3$ is:

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