JEE Challenger
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MathMedAdv 2017 • Paper 1

Matching Conic Tangent and Contact Point for Ellipse

Comprehension Passage

Answer Q.49, Q.50 and Q.51 by appropriately matching the details presented across the three columns in the table below.

Columns 1, 2, and 3 list various conics, the equations of their tangents, and their respective points of contact.

Column 1Column 2Column 3(I)x2+y2=a2(i)my=m2x+a(P)(am2,2am)(II)x2+a2y2=a2(ii)y=mx+am2+1(Q)(−mam2+1,am2+1)(III)y2=4ax(iii)y=mx+a2m2−1(R)(−a2ma2m2+1,1a2m2+1)(IV)x2−a2y2=a2(iv)y=mx+a2m2+1(S)(−a2ma2m2−1,−1a2m2−1)\begin{array}{|l|l|l|} \hline \text{Column 1} & \text{Column 2} & \text{Column 3} \\ \hline (\text{I}) \quad x^2 + y^2 = a^2 & (\text{i}) \quad my = m^2 x + a & (\text{P}) \quad \left(\frac{a}{m^2}, \frac{2a}{m}\right) \\ \hline (\text{II}) \quad x^2 + a^2 y^2 = a^2 & (\text{ii}) \quad y = mx + a\sqrt{m^2 + 1} & (\text{Q}) \quad \left(\frac{-ma}{\sqrt{m^2+1}}, \frac{a}{\sqrt{m^2+1}}\right) \\ \hline (\text{III}) \quad y^2 = 4ax & (\text{iii}) \quad y = mx + \sqrt{a^2 m^2 - 1} & (\text{R}) \quad \left(\frac{-a^2 m}{\sqrt{a^2 m^2 + 1}}, \frac{1}{\sqrt{a^2 m^2 + 1}}\right) \\ \hline (\text{IV}) \quad x^2 - a^2 y^2 = a^2 & (\text{iv}) \quad y = mx + \sqrt{a^2 m^2 + 1} & (\text{S}) \quad \left(\frac{-a^2 m}{\sqrt{a^2 m^2 - 1}}, \frac{-1}{\sqrt{a^2 m^2 - 1}}\right) \\ \hline \end{array}

Suppose the tangent line to a suitable conic in Column 1 at the point (3,12)(\sqrt{3}, \frac{1}{2}) is 3x+2y=4\sqrt{3}x + 2y = 4. Which among the given choices is the only CORRECT combination?

Options

A

(IV) (iii) (S)

B

(IV) (iv) (S)

C

(II) (iii) (R)

D

(II) (iv) (R)

Correct

Topics & Concepts

Conic SectionsEllipseTangents and Normals

Step-by-Step Solution

To find the correct combination, we need to test which conic from Column 1 passes through the given point (3,12)\left(\sqrt{3}, \frac{1}{2}\right) and has the tangent equation 3x+2y=4\sqrt{3}x + 2y = 4 at that point, and then determine its corresponding tangent form (Column 2) and point of contact (Column 3).

Step 1: Identify the Conic in Column 1

Consider Conic (II): x2+a2y2=a2orx2a2+y21=1x^2 + a^2 y^2 = a^2 \quad \text{or} \quad \frac{x^2}{a^2} + \frac{y^2}{1} = 1

Substitute the point (3,12)\left(\sqrt{3}, \frac{1}{2}\right) into the equation: (3)2+a2(12)2=a2(\sqrt{3})^2 + a^2 \left(\frac{1}{2}\right)^2 = a^2 3+a24=a2  ⟹  3a24=3  ⟹  a2=43 + \frac{a^2}{4} = a^2 \implies \frac{3a^2}{4} = 3 \implies a^2 = 4

Now, write the equation of the tangent to x2+4y2=4x^2 + 4y^2 = 4 at the point (3,12)\left(\sqrt{3}, \frac{1}{2}\right) using the T=0T = 0 formula: x(3)+4y(12)=4x(\sqrt{3}) + 4y\left(\frac{1}{2}\right) = 4 3x+2y=4\sqrt{3}x + 2y = 4

This matches the given tangent line 3x+2y=4\sqrt{3}x + 2y = 4. Thus, the correct conic in Column 1 is (II).

Step 2: Determine the Tangent Form in Column 2

For the ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 with b=1b = 1, the standard equation of a tangent with slope mm is given by: y=mx+a2m2+b2y = mx + \sqrt{a^2 m^2 + b^2}

Substituting b=1b = 1: y=mx+a2m2+1y = mx + \sqrt{a^2 m^2 + 1}

This matches item (iv) in Column 2.

Step 3: Determine the Point of Contact in Column 3

Let (x1,y1)(x_1, y_1) be the point of contact for the tangent y=mx+a2m2+1y = mx + \sqrt{a^2 m^2 + 1} on the ellipse x2a2+y2=1\frac{x^2}{a^2} + y^2 = 1.

The equation of the tangent at (x1,y1)(x_1, y_1) is: xx1a2+yy1=1  ⟹  y=−(x1a2y1)x+1y1\frac{x x_1}{a^2} + y y_1 = 1 \implies y = -\left(\frac{x_1}{a^2 y_1}\right)x + \frac{1}{y_1}

Comparing this with y=mx+a2m2+1y = mx + \sqrt{a^2 m^2 + 1}: 1y1=a2m2+1  ⟹  y1=1a2m2+1\frac{1}{y_1} = \sqrt{a^2 m^2 + 1} \implies y_1 = \frac{1}{\sqrt{a^2 m^2 + 1}}

m=−x1a2y1  ⟹  x1=−a2my1=−a2ma2m2+1m = -\frac{x_1}{a^2 y_1} \implies x_1 = -a^2 m y_1 = \frac{-a^2 m}{\sqrt{a^2 m^2 + 1}}

Thus, the point of contact is: (−a2ma2m2+1,1a2m2+1)\left(\frac{-a^2 m}{\sqrt{a^2 m^2 + 1}}, \frac{1}{\sqrt{a^2 m^2 + 1}}\right)

This matches item (R) in Column 3.

Conclusion

The correct combination is (II) (iv) (R), which corresponds to Option D.

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