Find the equation of one of the common tangen | vedclass

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(A) The equation of a tangent to the parabola $y^2 = 4x$ is $y = mx + \frac{1}{m}$.For this line to be a tangent to the ellipse $\frac{x^2}{4} + \frac

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$x + 2y + 4 = 0$

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(A) The equation of a tangent to the parabola $y^2 = 4x$ is $y = mx + \frac{1}{m}$.For this line to be a tangent to the ellipse $\frac{x^2}{4} + \frac{y^2}{3} = 1$,it must satisfy the condition $c^2 = a^2m^2 + b^2$,where $c = \frac{1}{m}$,$a^2 = 4$,and $b^2 = 3$.Substituting these values: $(\frac{1}{m})^2 = 4m^2 + 3$.Let $m^2 = t$,then $\frac{1}{t} = 4t + 3$,which implies $4t^2 + 3t - 1 = 0$.Factoring the quadratic: $(4t - 1)(t + 1) = 0$.Since $t = m^2$ must be positive,we have $t = \frac{1}{4}$,so $m = \pm \frac{1}{2}$.For $m = \frac{1}{2}$,the tangent is $y = \frac{1}{2}x + \frac{1}{1/2}$ $\Rightarrow y = \frac{1}{2}x + 2$ $\Rightarrow x - 2y + 4 = 0$.For $m = -\frac{1}{2}$,the tangent is $y = -\frac{1}{2}x - 2 \Rightarrow x + 2y + 4 = 0$.

Find the equation of one of the common tangents to the parabola $y^2 = 4x$ and the ellipse $\frac{x^2}{4} + \frac{y^2}{3} = 1$.

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