728 views

1 Answer

0 votes
0 votes

Answer: B

--------------------------------------------------------------

Let $P(t^2,2t)$ be the parametric point on the parabola. Then the slope of the parabola at $P$ is given by

$$\frac{dy}{dx}=\frac{1}{t}$$

Now, the slope of the tangent at $P$ passing through the point $P$ and $(1,4)$ is given by $$\frac{2t-4}{t^2-1}$$.

Hence, equating the slopes at point $P$, we have the following $$\frac{1}{t}=\frac{2t-4}{t^2-1}$$

The roots of the equation are $t_1$ and $t_2$.

The angle between the two tangents is given by

$$\tan\alpha =\left|\frac{t_2-t_1}{1+t_1t_2}\right|$$

On solving, we will get $\alpha=\frac{\pi}{3}$.

Related questions

1 votes
1 votes
1 answer
1
akash.dinkar12 asked May 11, 2019
947 views
Number of real solutions of the equation $x^7 + 2x^5 + 3x^3 + 4x = 2018$ is$1$$3$$5$$7$
2 votes
2 votes
1 answer
2
akash.dinkar12 asked May 11, 2019
717 views
The sum of the infinite series $1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\dots $ is$2$$3$$4$$6$
0 votes
0 votes
1 answer
3
akash.dinkar12 asked May 11, 2019
675 views
The $x$-axis divides the circle $x^2 + y^2 − 6x − 4y + 5 = 0$ into two parts. The area of the smaller part is$2\pi-1$$2(\pi-1)$$2\pi-3$$2(\pi-2)$
2 votes
2 votes
1 answer
4
akash.dinkar12 asked May 11, 2019
810 views
If $\alpha$ is a root of $x^2-x+1$, then $\alpha^{2018} + \alpha^{-2018}$ is$-1$$0$$1$$2$