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प्रश्न
Find the equation of the locus of a point the tangents form which to the ellipse 3x2 + 5y2 = 15 are at right angles
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उत्तर
The equation of the ellipse is 3x2 + 5y2 = 15
∴ `x^2/5 + y^2/3` = 1.
Comparing with `x^2/"a"^2 + y^2/"b"^2` = 1, we get,
a2 = 5, b2 = 3
Let P = (x1, y1) be any point on the required locus.
The equations of the tangents with slope m are
y = `"m"x ± sqrt("a"^2"m"^2 + "b"^2)`
i.e., y = `"m"x ± sqrt(5"m"^2 + 3)`
If these tangents pass through P(x1, y1), we have
y1 = `"m"x_1 ± sqrt(5"m"^2 + 3)`
∴ y1 – mx1 = `± sqrt(5"m"^2 + 3)`
Squaring both the sides, we get
(y1 – mx1)2 = 5m2 + 3
∴ `"y"_1^2 - 2"x"_1"y"_1"m" + "m"^2"x"_1^2 - 5"m"^2 - 3 = 0`
∴ `("x"_1^2 - 5)"m"^2 - 2"x"_1"y"_1"m" + ("y"_1^2 - 3) = 0`
This is a quadratic in m
Its roots m1 and m2 are the slopes of the tangents drawn from P.
From the quadratic equation,
m1m2 = `(y_1^2 - 3)/(x_1^2 - 5)`
But the tangents from P are at right angles
∴ m1m2 = – 1
∴ `(y_1^2 - 3)/(x_1^2 - 5)` = – 1
∴ `"y"_1^2 - 3 = -"x"_1^2 + 5`
∴ `"x"_1^2 + "y"_1^2 = 8`
Replacing x1 by x and y1 by y, the equation of required locus is x2 + y2 = 8.
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