Question

Let ${\rm P}$ be a point on the parabola, ${x^2} = 4y$ If the distance of ${\rm P}$ from the centre of the circle, ${x^2} + {y^2} + 6x + 8 = 0$ is minimum, then the equation of the tangent to the parabola at ${\rm P}$ , is?
A. $x + 4y - 2 = 0$
B. $x + 2y = 0$
C. $x + y + 1 = 0$
D. $x - y + 3 = 0$

Answer 1

Hint : In order to determine the equation of the tangent to the parabola at ${\rm P}$ and the line drawn from the centre of the circle to point ${\rm P}$ must be natural to the parabola ${x^2} = 4y$ at the point ${\rm P}$ in order for the distance between point ${\rm P}$ $\left( { - g, - f} \right)$ and the centre of the circle of the equation ${x^2} + {y^2} + 6x + 8 = 0$ to be as small as possible. we use the tangent formula $y - {y_1} = m(x - {x_1})$ with the point $P({x_1},{y_1})$ to find the required answer.

Complete step-by-step answer :
We are given the ${\rm P}$ be a point on the parabola, ${x^2} = 4y$ , centre of the circle, ${x^2} + {y^2} + 6x + 8 = 0$ is minimum.
We need to find out the equation of the tangent to the parabola at ${\rm P}$
Let the point ${\rm P}$ on parabola be $(2t,{t^2})$ and the centre $\left( { - g, - f} \right)$ .Then the centre will be: $\left( { - 3,{\text{ }}0} \right)$
Now slope for line , $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ from centre to point $({x_1},{y_1}) = ( - 3,0)$ and $({x_2},{y_2}) = (2t,{t^2})$

m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \dfrac{{{t^2} - 0}}{{2t + 3}} \\\ m = \dfrac{{{t^2}}}{{2t + 3}} \to (1) \; \ $$ To find the slope of a line tangent to a parabola at a specific point, find the derivative of the parabola's equation, $${x^2} = 4y$$ By differentiating the parabola equation with respect to x, we get $$\ 2xdx = 4dy \\\ \dfrac{{2x}}{4} = \dfrac{{dy}}{{dx}} \\\ \dfrac{{dy}}{{dx}} = \dfrac{x}{2} \; \ $$ So, the slope of tangent to the parabola will be : $$m = \dfrac{x}{2} = t$$ Slope of normal is: $$ - \dfrac{1}{m} = - \dfrac{1}{t} \to (2)$$ Now, comparing the equation $$(1)$$ and $$(2)$$ , we can get $$\dfrac{{{t^2}}}{{2t + 3}} = - \dfrac{1}{t}$$ Expanding the equation on RHS, we can get $$\dfrac{{{t^2}}}{{2t + 3}} + \dfrac{1}{t} = 0$$ Take LCM on both side, we get $$\ \dfrac{{t({t^2}) + 2t + 3}}{{t(2t + 3)}} = 0 \\\ {t^3} + 2t + 3 = 0 \; \ $$ In this case we find that by substituting $$t$$ with $$ - 1$$ , it actually satisfies the equation, so $$ - 1$$ is one of its roots. After that, divide the equation by $$t + 1$$ (since $$ - 1$$ is one of the roots) using long division. Now, we have $${t^3} + 2t + 3 = (t + 1)({t^2} - t + 1)$$ With this we get a quadratic equation and by factoring it or using the quadratic formula, we can find the other 2 roots. $$(t + 1)({t^2} - t + 1) = 0$$ So, the real root will be $$(t + 1)$$ : $$t = - 1$$ and the coordinates of $$P(2t,{t^2})$$ $$(2t,{t^2}) = (2( - 1),{( - 1)^2}) = ( - 2,1)$$ . Now, the slope of tangent to parabola at P will be: $$m = \dfrac{x}{2} = t = - 1$$ So the equation of tangent will be: $$y - {y_1} = m(x - {x_1})$$ with the point $$({x_1},{y_1}) = ( - 2,1)$$ $$\ y - 1 = - 1(x - ( - 2)) \\\ y - 1 = - 1(x + 2) \\\ y - 1 + x + 2 = 0 \\\ x + y + 1 = 0 \; \ $$ Finally, the point $${\rm P}$$ on the parabola, $${x^2} = 4y$$ If the distance of $${\rm P}$$ from the centre of the circle, $${x^2} + {y^2} + 6x + 8 = 0$$ is minimum, then the equation of the tangent to the parabola at $${\rm P}$$ , is $$x + y + 1 = 0$$ . Hence, the option( C ) $$x + y + 1 = 0$$ is the correct answer. **So, the correct answer is “Option C”.** **Note** : First, we plot a graph is shown below The green curve represent the parabola equation $${x^2} = 4y$$ The blue line represent the tangent equation $$x + y + 1 = 0$$ The centre of the circle with the point $$\left( { - 3,{\text{ }}0} \right)$$ of the equation $${x^2} + {y^2} + 6x + 8 = 0$$  We use the formula to find the slope of the line, $$m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$$ and $$y - {y_1} = m(x - {x_1})$$ is the formula for find out the equation of tangent.