Question

Let P be a point on the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ and let the perpendicular drawn through P to the major axis meet its auxiliary circle at Q. If the normals drawn at P and Q to the ellipse and the auxiliary circle respectively meet in R, then the equation of the locus of R is

• A. $x^2+y^2=5$
• B. $x^2+y^2=13$
• C. $x^2+y^2=25$
• D. $x^2+y^2=1$

Hint:

This problem has a standard result: the locus of the intersection of normals to an ellipse and its auxiliary circle at corresponding points P and Q is the circle $x^2+y^2=(a+b)^2$. Here, $a=3, b=2$, so the locus is $x^2+y^2=(3+2)^2=25$. Knowing this result can lead to the answer instantly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
This problem requires finding the locus of the intersection of two normal lines, one to an ellipse and one to its auxiliary circle, where the points of normalcy are related. We will use the parametric equations for the normals and eliminate the parameter to find the locus.
Step 2: Key Formula or Approach
1. Identify the parameters $a$ and $b$ for the ellipse. The auxiliary circle is $x^2+y^2=a^2$. 2. Represent a general point P on the ellipse parametrically as $(a\cos\phi, b\sin\phi)$. 3. Find the coordinates of the corresponding point Q on the auxiliary circle. 4. Write the equation of the normal to the ellipse at P. 5. Write the equation of the normal to the auxiliary circle at Q. 6. Solve these two equations simultaneously to find the coordinates of their intersection point R in terms of the parameter $\phi$. Then eliminate $\phi$ to find the locus.
Step 3: Detailed Explanation
1. Parameters and parametric points: The ellipse is $\frac{x^2}{9}+\frac{y^2}{4}=1$. So, $a^2=9 \implies a=3$ and $b^2=4 \implies b=2$. The auxiliary circle is $x^2+y^2=a^2=9$. Let P be a point on the ellipse with eccentric angle $\phi$. The coordinates of P are $(3\cos\phi, 2\sin\phi)$. The perpendicular from P to the major axis (the x-axis) is the line $x=3\cos\phi$. This line intersects the auxiliary circle $x^2+y^2=9$ at the point Q. Substituting $x=3\cos\phi$ gives $(3\cos\phi)^2+y^2=9 \implies 9\cos^2\phi+y^2=9 \implies y^2=9(1-\cos^2\phi)=9\sin^2\phi \implies y=\pm 3\sin\phi$. The point Q corresponding to P has the same eccentric angle, so its coordinates are $(3\cos\phi, 3\sin\phi)$. 2. Equation of normal to the ellipse at P: The standard equation of the normal to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $(a\cos\phi, b\sin\phi)$ is $ax\sec\phi - by\csc\phi = a^2-b^2$. For our ellipse, this becomes: \[ 3x\sec\phi - 2y\csc\phi = 9-4 = 5 \] \[ \frac{3x}{\cos\phi} - \frac{2y}{\sin\phi} = 5 \quad \text{(Equation 1)} \] 3. Equation of normal to the auxiliary circle at Q: The normal to any circle passes through its center. The auxiliary circle $x^2+y^2=9$ has its center at the origin (0,0). So, the normal at Q$(3\cos\phi, 3\sin\phi)$ is the line passing through Q and the origin. Its equation is $y = \left(\frac{3\sin\phi}{3\cos\phi}\right)x \implies y = (\tan\phi)x$. 4. Find the locus of the intersection point R(h,k): Let R(h,k) be the point of intersection. Its coordinates must satisfy both normal equations. From the circle's normal equation: $k = (\tan\phi)h \implies \frac{k}{h} = \tan\phi = \frac{\sin\phi}{\cos\phi}$. This means $\sin\phi$ is proportional to $k$ and $\cos\phi$ is proportional to $h$. Let $\sin\phi = k\lambda$ and $\cos\phi = h\lambda$. Using $\cos^2\phi+\sin^2\phi=1$, we get $(h\lambda)^2+(k\lambda)^2=1 \implies \lambda^2(h^2+k^2)=1 \implies \lambda = \frac{1}{\sqrt{h^2+k^2}}$. So, $\cos\phi = \frac{h}{\sqrt{h^2+k^2}}$ and $\sin\phi = \frac{k}{\sqrt{h^2+k^2}}$. Now substitute these into the ellipse's normal equation (Equation 1) with $(x,y)=(h,k)$: \[ \frac{3h}{\cos\phi} - \frac{2k}{\sin\phi} = 5 \] \[ \frac{3h}{h/\sqrt{h^2+k^2}} - \frac{2k}{k/\sqrt{h^2+k^2}} = 5 \] \[ 3\sqrt{h^2+k^2} - 2\sqrt{h^2+k^2} = 5 \] \[ \sqrt{h^2+k^2} = 5 \] Squaring both sides: \[ h^2+k^2 = 25 \] Replacing $(h,k)$ with $(x,y)$ gives the locus. Step 4: Final Answer
The equation of the locus of R is $x^2+y^2=25$.