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Tangents are drawn from a point P to the hyperbola `x^2-y^2= a^2` If the chord of contact of these normal to the curve, prove that the locus of P is `1/x^2 - 1/y^2 = 4/a^2` |
| Answer» Let locus of point `P(h,k)``Eqn.` of tangent from point `P` to the hyperbola is:`y = mx pm sqrt(a^2*m^2 - a^2)``y = mx pm a*sqrt(m^2 -1)``Eqn.` of normal to this hyperbola:`a^2*y_1(x-x_1) + a^2*x_1(y-y_1)`Tangents starts from point `P(h,k)` , so it is point of contact. And since this point is normal to the curve, it satisfies `Eqn.` of normal.`a^2*k(x-h) + a^2*h(y-k) = 0`Converting it to slope form:`kx - kh + hy - kh = 0``y = -k/h*x + 2k`And from equation of tangent:`k = mh pm a*sqrt(m^2-1)``rArr m = -k/h` ......`(eq.1)`and,`a*sqrt(m^2-1) = 2k`Squaring both sides`a^2*(m^2-1) = 4k^2``a^2*((-k/h)^2 - 1) = 4k^2``k^2/h^2 - 1 = 4k^2/a^2`Divinding above `Eqn.` by `k^2``1/h^2 -1/k^2 = 4/a^2``:.` locus of point P(h,k) is`1/x^2 -1/y^2 = 4/a^2` | |