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# If the angle between the tangents drawn from a point $P$ to the parabola ${{y}^{2}}=4ax$ is ${{45}^{\circ }}$, then the locus of $P$ isa. Parabolab. Ellipsec. Hyperbolad. Circle

Last updated date: 12th Sep 2024
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Hint: To find the locus of point from which tangents to the parabola are drawn at a certain angle, write the equation of tangents at any two points on the parabola and find their point of intersection. Use the angle formula to find the relation between the slope of the two tangents.

Complete step-by-step answer:
We have a parabola ${{y}^{2}}=4ax$ to which two tangents are drawn from a point and the angle between those two tangents is ${{45}^{\circ }}$.
Let’s assume that the tangents drawn from point $P$ touch the parabola ${{y}^{2}}= 4ax$ at points$Q\left( {{t}_{1}} \right)=\left( at_{1}^{2},2a{{t}_{1}} \right)$and$R\left( {{t}_{2}} \right)=\left( at_{2}^{2},2a{{t}_{2}} \right)$.
We know that the equation of tangents at these two points will intersect at $P$ whose coordinates are of the form$\left[ a{{t}_{1}}{{t}_{2}},a\left( {{t}_{1}}+{{t}_{2}} \right) \right]$.
The equation of tangent at any point$\left( a{{t}^{2}},2at \right)$ of the parabola ${{y}^{2}}=4ax$ is of the form $y=\dfrac{x}{t}+at$
So, the slope of tangent through $Q\left( {{t}_{1}} \right)$ is $\dfrac{1}{{{t}_{1}}}$ and $R\left( {{t}_{2}} \right)$ is $\dfrac{1}{{{t}_{2}}}$.
We know that the angle $\alpha$ between two lines of slope ${{m}_{1}}$ and ${{m}_{2}}$ has the formula $\tan \alpha =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$
So, the angle between tangents of slope $\dfrac{1}{{{t}_{1}}}$ and $\dfrac{1}{{{t}_{2}}}$ is ${{45}^{\circ }}$
$\Rightarrow \tan {{45}^{\circ }}=1=\left| \dfrac{\dfrac{1}{{{t}_{1}}}-\dfrac{1}{{{t}_{2}}}}{1+\dfrac{1}{{{t}_{1}}}\cdot \dfrac{1}{{{t}_{2}}}} \right|$
$\Rightarrow \dfrac{1}{{{t}_{1}}}-\dfrac{1}{{{t}_{2}}}=\pm \left( 1+\dfrac{1}{{{t}_{1}}}\cdot \dfrac{1}{{{t}_{2}}} \right)$
$\Rightarrow {{t}_{2}}-{{t}_{1}}=\pm \left( {{t}_{1}}{{t}_{2}}+1 \right)$
To find the value of ${{t}_{1}}+{{t}_{2}}$, we use the formula ${{t}_{1}}+{{t}_{2}}=\sqrt{{{\left( {{t}_{2}}-{{t}_{1}} \right)}^{2}}+4{{t}_{1}}{{t}_{2}}}$
\begin{align} & \Rightarrow {{t}_{1}}+{{t}_{2}}=\sqrt{{{\left( {{t}_{1}}{{t}_{2}}+1 \right)}^{2}}+4{{t}_{1}}{{t}_{2}}} \\ & \Rightarrow {{\left( {{t}_{1}}+{{t}_{2}} \right)}^{2}}=1+6{{t}_{1}}{{t}_{2}}+{{\left( {{t}_{1}}{{t}_{2}} \right)}^{2}} \\ & \\ \end{align} $(1)$
We know coordinates of $P$ are of the form $\left[ a{{t}_{1}}{{t}_{2}},a\left( {{t}_{1}}+{{t}_{2}} \right) \right]$.
Let’s assume $x=a{{t}_{1}}{{t}_{2}},y=a\left( {{t}_{1}}+{{t}_{2}} \right)$
$\Rightarrow \dfrac{x}{a}={{t}_{1}}{{t}_{2}},\dfrac{y}{a}=\left( {{t}_{1}}+{{t}_{2}} \right)$
Substituting these values in equation$(1)$, we get${{\left( \dfrac{y}{a} \right)}^{2}}=1+6\dfrac{x}{a}+{{\left( \dfrac{x}{a} \right)}^{2}}$
Adding $9$ on both sides of above equation, we get${{\left( \dfrac{y}{a} \right)}^{2}}+9=1+9+6\dfrac{x}{a}+{{\left( \dfrac{x}{a} \right)}^{2}}$
\begin{align} & \Rightarrow {{\left( \dfrac{y}{a} \right)}^{2}}+9=1+{{\left( \dfrac{x}{a}+3 \right)}^{2}} \\ & \Rightarrow {{\left( \dfrac{x}{a}+3 \right)}^{2}}-{{\left( \dfrac{y}{a} \right)}^{2}}=8 \\ & \Rightarrow \dfrac{{{\left( \dfrac{x}{a}+3 \right)}^{2}}}{8}-\dfrac{{{\left( \dfrac{y}{a} \right)}^{2}}}{8}=1 \\ \end{align}
We observe that the locus of points from which two tangents are drawn at a certain angle is hyperbola.
However, it is not necessary that we will always get a hyperbola. The locus of curve changes with change in angle between the two tangents. If the two tangents are perpendicular to each other, we will get the equation of locus of their point of intersection as a straight line.
Hence, the correct answer is (c) Hyperbola.

Note: We can also find the locus of $P$ by taking any general point instead of using parametric form and then write the equation of tangents from the general point. Keep in mind that while removing the modulus, we will consider both negative and positive values.