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Question: Find the length of the chord of contact of the tangents drawn from the point \(\left( {{x}_{1}},{{y}...

Find the length of the chord of contact of the tangents drawn from the point (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) to the parabola y2=4ax{{y}^{2}}=4ax

Explanation

Solution

We need to find the length of the tangents to the parabola y2=4ax{{y}^{2}}=4ax . Let us consider that the tangents at points P and Q meet at (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) . We start to solve the question by finding out the value of the coordinates (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) . Then, we find the distance between the points P, Q to get the desired result.

Complete step-by-step answer:
We are given an equation of a parabola and need to find the length of chord of contact of the tangents. We start to solve the question by finding out the value of coordinates (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) and then find the length of the chord of contact of the tangents.
A line touching a parabola is said to be a tangent to a parabola. The tangent to a parabola touches it at exactly one point.
The chord that joins the point of contact of the tangents from an external point is called the chord of contact of a parabola.
The given question can be diagrammatically represented as follows,

From the above figure,
The tangents to the parabola meet at the points A,B  A,B\; on the parabola.
The points A,B  A,B\; are given as follows,
A=(at12,2at1)A=\left( a{{t}_{1}}^{2},2a{{t}_{1}} \right) ;
B=(at22,2at2)B=\left( a{{t}_{2}}^{2},2a{{t}_{2}} \right)
The tangents at the points A,B  A,B\; meets at a point (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right)
From the above,
The coordinates of (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) are given as follows,
x1=at1t2{{x}_{1}}=a{{t}_{1}}{{t}_{2}}
y1=a(t1+t2){{y}_{1}}=a\left( {{t}_{1}}+{{t}_{2}} \right)
The length of the chord of contact of the tangents drawn from the point (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) to the parabola y2=4ax{{y}^{2}}=4ax is the distance between the points A,B  A,B\; .
We know that distance between two points (x1,y1),(x2,y2)\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right) is given by the formula
=(x1x2)2+(y1y2)2= \sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}
Following the same, the distance between the points A(at12,2at1),B(at22,2at2)A\left( a{{t}_{1}}^{2},2a{{t}_{1}} \right),B\left( a{{t}_{2}}^{2},2a{{t}_{2}} \right) is given by
=(at12at22)2+(2at12at2)2= \sqrt{{{\left( a{{t}_{1}}^{2}-a{{t}_{2}}^{2} \right)}^{2}}+{{\left( 2a{{t}_{1}}-2a{{t}_{2}} \right)}^{2}}}
Simplifying the above equation, we get,
=a2(t12t22)2+4a2(t1t2)2= \sqrt{{{a}^{2}}{{\left( {{t}_{1}}^{2}-{{t}_{2}}^{2} \right)}^{2}}+4{{a}^{2}}{{\left( {{t}_{1}}-{{t}_{2}} \right)}^{2}}}
Taking a2{{a}^{2}} from the square root, we get,
=a(t12t22)2+2(t1t2)2= a\sqrt{{{\left( {{t}_{1}}^{2}-{{t}_{2}}^{2} \right)}^{2}}+2{{\left( {{t}_{1}}-{{t}_{2}} \right)}^{2}}}
Simplifying the above equation by expanding the terms, we get,
=a((t1+t2)24t1t2)((t1+t2)2+4)= a\sqrt{\left( {{\left( {{t}_{1}}+{{t}_{2}} \right)}^{2}}-4{{t}_{1}}{{t}_{2}} \right)\left( {{\left( {{t}_{1}}+{{t}_{2}} \right)}^{2}}+4 \right)}
Substituting the values with x1,y1{{x}_{1,}}{{y}_{1}} in the above equation, we get,
=a((y1a)24(x1a))((y1a)2+4)= a\sqrt{\left( {{\left( \dfrac{{{y}_{1}}}{a} \right)}^{2}}-4\left( \dfrac{{{x}_{1}}}{a} \right) \right)\left( {{\left( \dfrac{{{y}_{1}}}{a} \right)}^{2}}+4 \right)}
Simplifying the above equation, we get,
=a(y124ax1a2)(y12+4a2a2)= a\sqrt{\left( \dfrac{{{y}_{1}}^{2}-4a{{x}_{1}}}{{{a}^{2}}} \right)\left( \dfrac{{{y}_{1}}^{2}+4{{a}^{2}}}{{{a}^{2}}} \right)}
Taking a2{{a}^{2}} out of the square root, we get,
=aa(y124ax1)(y12+4a2)a2= \dfrac{a}{a}\sqrt{\dfrac{\left( {{y}_{1}}^{2}-4a{{x}_{1}} \right)\left( {{y}_{1}}^{2}+4{{a}^{2}} \right)}{{{a}^{2}}}}
=(y124ax1)(y12+4a2)a2= \sqrt{\dfrac{\left( {{y}_{1}}^{2}-4a{{x}_{1}} \right)\left( {{y}_{1}}^{2}+4{{a}^{2}} \right)}{{{a}^{2}}}}

Note: We need to know that distance between two points (x1,y1),(x2,y2)\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right) is given by the formula
=(x1x2)2+(y1y2)2= \sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}
The values of the coordinates (x1,y1)\left( {{x}_{1}},{{y}_{1}} \right) on the parabola y2=4ax{{y}^{2}}=4ax is given by x1=at1t2{{x}_{1}}=a{{t}_{1}}{{t}_{2}}
y1=a(t1+t2){{y}_{1}}=a\left( {{t}_{1}}+{{t}_{2}} \right) such that they are the point of intersection of both the tangents to the parabola.