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Two tangents are drawn from the point $(-2,-1)$ to the parabola $y^2=4x$. If $\alpha$ is the angle between them then find the value of $\tan \alpha$ .

My try:

Eqn of tangent $1,T_1$ say at the point $(x_1,y_1)$ is $yy_1=2a(x+x_1)$, Eqn of tangent $2,T_2$ say at the point $(x_2,y_2)$ is $yy_2=2a(x+x_2)$

Since they both pass through $(-2,-1)$ hence we have $y_1=2(2-x_1),y_2=2(2-x_2)$.

Angle between two tangents is $\tan \alpha=\vert\dfrac{m_1-m_2}{1+m_1m_2}\lvert$

Now $m_1=\dfrac{y_1+1}{x_1+2},m_2=\dfrac{y_2+1}{x_2+1}$

Replacing $y_1,y_2$ is not giving the required angle.

How to do it?Please help.

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  • $\begingroup$ The denominator of $m_2$ should be $x_2+2$. If that’s just a typo, you’ll need to show your solutions for the two intersections of the tangents with the parabola for us to determine where you went wrong. $\endgroup$ – amd Jan 11 '18 at 8:22
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Any point on $y^2=4x$ P$(t^2,2t)$

$$\dfrac{dy}{dx}_{(\text{ at }P)}=\dfrac4{2y}_{(\text{ at }P)}=\dfrac1t$$

$$\implies\dfrac1t=\dfrac{2t+1}{t^2+2}\iff t^2+t-2=0$$

If $t_1.t_2$ are the roots of the above equation, WLOG $t_1=1,t_2=-2$

So, $\tan\alpha=\left|\dfrac{\dfrac1{t_1}-\dfrac1{t_2}}{1+\dfrac1{t_1}\cdot\dfrac1{t_2}}\right|=\left|\dfrac{t_2-t_1}{t_1t_2+1}\right|$

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Taking the derivative wrt $x$, $y^2=4x$ becomes $2yy'= 4$, or $y' = \dfrac{2}{y}$.

So, at the point $(\frac 14t^2,t)$ on the parabola, the slope of the tangent is $m = \dfrac 2t$.

So the equation of the tangent line through the point $(\frac 14t^2,t)$ is $y-t = \frac 2t(x-\frac 14t^2)$.

For what values of $t$ does this line pass through the point $(-2,-1)$?

\begin{align} -1-t &= \frac 2t(-2-\frac 14t^2) \\ 1+t &= \frac 4t + \frac t2\\ 2t + 2t^2 &= 8 + t^2\\ t^2+2t-8 &= 0 \\ (t+4)(t-2) &= 0 \\ t &\in \{2, -4\} \\ (x,y) &\in \{(1,2), (4,-4)\} \\ m &\in \left\{1, -\frac 12 \right\} \end{align}

\begin{align} \tan\alpha &=\left|\dfrac{m_1-m_2}{1+m_1 m_2}\right| \\ &=\left|\dfrac{1+\frac 12}{1-\frac 12}\right| \\ &= 3 \end{align}

$$\alpha \approx 71.57^\circ$$

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