The equation of the tangent to the parabola y | vedclass

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Question

(D) The equation of the parabola is $y^2 = 12x$,which is of the form $y^2 = 4ax$,where $4a = 12$,so $a = 3$. Let the point of contact be $(x_1, y_1)$.

Vedclass · Accepted Answer

$(9, 6\sqrt{3})$

Vedclass · Answer

(D) The equation of the parabola is $y^2 = 12x$,which is of the form $y^2 = 4ax$,where $4a = 12$,so $a = 3$. Let the point of contact be $(x_1, y_1)$. The equation of the tangent to the parabola $y^2 = 4ax$ at $(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$. Substituting $a = 3$,we get $yy_1 = 6(x + x_1)$,which simplifies to $6x - y_1y + 6x_1 = 0$. We are given the tangent equation as $x - \sqrt{3}y + 9 = 0$. Comparing the two equations: $\frac{6}{1} = \frac{-y_1}{-\sqrt{3}} = \frac{6x_1}{9}$. From $\frac{6}{1} = \frac{y_1}{\sqrt{3}}$,we get $y_1 = 6\sqrt{3}$. From $\frac{6}{1} = \frac{6x_1}{9}$,we get $x_1 = 9$. Thus,the point of contact is $(9, 6\sqrt{3})$.

The equation of the tangent to the parabola $y^2=12x$,which makes an angle $30^{\circ}$ with the positive direction of the $X$-axis is given by $x-\sqrt{3}y+9=0$. The point of contact is:

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