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Let the transverse axis ofa varying hype...

Let the transverse axis ofa varying hyperbola be fixed with length of transverse axis being 2a. Then the locus of the point of contact of any tangent drawn to it from a fixed point on conjugate axis is

A

a parabola

B

a circle

C

an ellipse

D

a hyperbola

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To solve the problem step by step, we need to find the locus of the point of contact of any tangent drawn to a hyperbola from a fixed point on the conjugate axis. ### Step 1: Understand the Hyperbola The standard form of a hyperbola with a transverse axis along the x-axis is given by: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] where the length of the transverse axis is \(2a\). ### Step 2: Equation of the Tangent The equation of the tangent to the hyperbola at a point \((x_0, y_0)\) on the hyperbola can be expressed as: \[ \frac{x}{a} \sec \phi - \frac{y}{b} \tan \phi = 1 \] where \(\phi\) is the angle corresponding to the point of tangency. ### Step 3: Identify the Fixed Point on the Conjugate Axis Let the fixed point on the conjugate axis be \(P(0, -b \cot \phi)\). This point is on the y-axis, where the conjugate axis is vertical. ### Step 4: Find the Point of Contact The point of contact \(Q\) of the tangent can be represented as: \[ Q(a \sec \phi, b \tan \phi) \] ### Step 5: Determine the Intersection with the Y-axis The tangent intersects the y-axis at point \(P\). The coordinates of \(P\) can be expressed as: \[ P(0, -b \cot \phi) \] ### Step 6: Relate the Variables Since \(P\) is fixed, we can set: \[ -b \cot \phi = \lambda \] where \(\lambda\) is a constant. ### Step 7: Substitute and Eliminate Variables Now, substituting \(b \cot \phi\) in the tangent equation and eliminating \(\phi\) and \(b\) gives: \[ \frac{x^2}{a^2} - \frac{y^2}{\lambda^2} = 1 \] ### Step 8: Identify the Locus Rearranging the equation leads us to: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] This is the equation of a parabola. ### Final Answer Thus, the locus of the point of contact of any tangent drawn to the hyperbola from a fixed point on the conjugate axis is a **parabola**. ---

To solve the problem step by step, we need to find the locus of the point of contact of any tangent drawn to a hyperbola from a fixed point on the conjugate axis. ### Step 1: Understand the Hyperbola The standard form of a hyperbola with a transverse axis along the x-axis is given by: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] where the length of the transverse axis is \(2a\). ...
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