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Question: Find the locus of point P such that chord of contact on point P with respect to \[{y^2} = 4ax\] touc...

Find the locus of point P such that chord of contact on point P with respect to y2=4ax{y^2} = 4ax touches the hyperbola x2y2=a2{x^2} - {y^2} = {a^2}.

Explanation

Solution

The chord which intersects two tangents drawn from two points from the parabola to a point outside the circle is known as chord of contact. For this particular question we calculate the equation of chord of contact and the equation of tangent of circle. These two equations are then compared to find the locus of point P.

Complete step by step solution:
Let point P be (x1,y1)({x_1},{y_1}).
We know that the equation of chord of contact on point P with respect to y2=4ax{y^2} = 4ax is
yy1=2a(x+x1) y=2axy1+2ax1y1(i)  y{y_1} = 2a\left( {x + {x_1}} \right) \\\ y = \dfrac{{2ax}}{{{y_1}}} + \dfrac{{2a{x_1}}}{{{y_1}}} - (i) \\\
Hence, the slope of the line is
m=2ay1(ii)m = \dfrac{{2a}}{{{y_1}}} - (ii)
We also know that equation of a tangent of hyperbola x2a2y2b2=1\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1 is
y=mx±a2m2b2(iii)y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} - (iii)
When we compare this equation with the hyperbola x2y2=a2{x^2} - {y^2} = {a^2}, the we see that
a2=b2 a=b(iv)  {a^2} = {b^2} \\\ a = b - (iv) \\\
Now, we compare (i)(i) and (iii)(iii) equation,
2axy1+2ax1y1=mx±a2m2b2\dfrac{{2ax}}{{{y_1}}} + \dfrac{{2a{x_1}}}{{{y_1}}} = mx \pm \sqrt {{a^2}{m^2} - {b^2}}
Substituting the value of mm from (ii)(ii), we get
2axy1+2ax1y1=2ay1x±a2(2a)(y1)22b2 2axy1+2ax1y1=2axy1±a2(2a)(y1)22b2 2ax1y1=a2(2a)(y1)22b2  \dfrac{{2ax}}{{{y_1}}} + \dfrac{{2a{x_1}}}{{{y_1}}} = \dfrac{{2a}}{{{y_1}}}x \pm \sqrt {{a^2}{{\dfrac{{(2a)}}{{{{({y_1})}^2}}}}^2} - {b^2}} \\\ \dfrac{{2a{x}}}{{{y_1}}} + \dfrac{{2a{x_1}}}{{{y_1}}} = \dfrac{{2a{x}}}{{{y_1}}} \pm \sqrt {{a^2}{{\dfrac{{(2a)}}{{{{({y_1})}^2}}}}^2} - {b^2}} \\\ \dfrac{{2a{x_1}}}{{{y_1}}} = \sqrt {{a^2}{{\dfrac{{(2a)}}{{{{({y_1})}^2}}}}^2} - {b^2}} \\\
Squaring both sides we get
(2ax1y1)2=(a2(2a)(y1)22b2)2 4a2x12y12=4a4y12b2y12  {\left( {\dfrac{{2a{x_1}}}{{{y_1}}}} \right)^2} = {\left( {\sqrt {{a^2}{{\dfrac{{(2a)}}{{{{({y_1})}^2}}}}^2} - {b^2}} } \right)^2} \\\ \dfrac{{4{a^2}x_1^2}}{{{y}_1^2}} = \dfrac{{4{a^4} - y_1^2{b^2}}}{{{y}_1^2}} \\\
Substituting the value from (iv)

4a2x12=4a4y12a2 4a2x12=a2(4a2y12) 4x12+y12=4a2 4{a^2}x_1^2 = 4{a^4} - y_1^2{a^2} \\\ 4{{{a}}^2}x_1^2 = {{{a}}^2}(4{a^2} - y_1^2) \\\ 4x_1^2 + y_1^2 = 4{a^2} \\\

Hence, the locus of the point P is 4x2+y2=4a24x_{}^2 + y_{}^2 = 4{a^2} which represents an ellipse.

Note:
We can also calculate the length of the chord of contact by drawing a perpendicular from the centre to the tangent and calculating the perpendicular distance between the lines.