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Locus of P When Tangent Axis-Intercepts Are Concyclic | JEE

JEE Maths question with a full step-by-step solution.

Question
Two tangents are drawn to the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 from the point P(h,k)P(h,k). The points in which these tangents cut the axes are concyclic. The locus of PP is:
Axy=a2b2xy=a^2b^2
Bx2a2−y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1
Cx2−y2=a2−b2x^2-y^2=a^2-b^2correct
Dx2+y2=a2+b2x^2+y^2=a^2+b^2
Solution
Step 1: The pair of tangents from P(h,k)P(h,k) is SS1=T2SS_1=T^2, i.e.
(x2a2+y2b2−1)(h2a2+k2b2−1)=(hxa2+kyb2−1)2.\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}-1\right)\left(\frac{h^2}{a^2}+\frac{k^2}{b^2}-1\right)=\left(\frac{hx}{a^2}+\frac{ky}{b^2}-1\right)^2.
 Step 2: Four points (two on the xx-axis, two on the yy-axis) are concyclic precisely when the power of the origin matches on both axes, i.e. x1x2=y1y2x_1x_2=y_1y_2. Step 3: Putting y=0y=0 and x=0x=0 in the pair of tangents and forming the products of roots gives
x1x2=−h2b2+k2a2k2−b2,y1y2=−h2b2+k2a2h2−a2.x_1x_2=-\frac{h^2b^2+k^2a^2}{k^2-b^2},\qquad y_1y_2=-\frac{h^2b^2+k^2a^2}{h^2-a^2}.
 Step 4: Setting x1x2=y1y2x_1x_2=y_1y_2,
k2−b2=h2−a2  ⇒  h2−k2=a2−b2.k^2-b^2=h^2-a^2 \;\Rightarrow\; h^2-k^2=a^2-b^2.
 ∴\therefore the locus is x2−y2=a2−b2x^2-y^2=a^2-b^2. Correct answer: (3)
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