Question:
If the distance between the foci of an ellipse
\[
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
\]
is \(6\) and the distance between its directrices is \(10\), then the equation of one of the tangents of the ellipse drawn parallel to the line
\[
y=\sqrt2\,x+5
\]
is
If the distance between the foci of an ellipse
\[
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
\]
is \(6\) and the distance between its directrices is \(10\), then the equation of one of the tangents of the ellipse drawn parallel to the line
\[
y=\sqrt2\,x+5
\]
is
Show Hint
For the ellipse
\[
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,
\]
a tangent with slope \(m\) is
\[
y=mx\pm\sqrt{a^2m^2+b^2}.
\]
Once \(a^2\) and \(b^2\) are known, the tangent can be written immediately.
Updated On: Jul 9, 2026
- \[ y=\sqrt2\,x+\sqrt{66} \]
- \[ y=\sqrt2\,x+12 \]
- \[ y=\sqrt2\,x+\sqrt{44} \]
- \[ y=\sqrt2\,x+6 \] \bigskip
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
For the ellipse
\[
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,
\]
\[
c^2=a^2-b^2,
\qquad
e=\frac{c}{a}.
\]
The directrices are
\[
x=\pm\frac{a}{e}.
\]
A tangent to the ellipse having slope \(m\) is
\[
y=mx\pm\sqrt{a^2m^2+b^2}.
\]
Step 1: Use the distance between the foci. Distance between the foci is \[ 2c=6. \] Hence, \[ c=3. \] \[ c^2=9. \] \[ \cdots (1) \]
Step 2: Use the distance between the directrices. The directrices are \[ x=\pm\frac{a}{e}. \] Therefore, \[ 2\frac{a}{e}=10. \] \[ \frac{a}{e}=5. \] Since \[ e=\frac{c}{a}, \] \[ \frac{a^2}{c}=5. \] Using \[ c=3, \] \[ a^2=15. \] \[ \cdots (2) \]
Step 3: Find \(b^2\). Using \[ c^2=a^2-b^2, \] \[ 9=15-b^2. \] \[ b^2=6. \]
Step 4: Find the tangent parallel to \(y=\sqrt2\,x+5\). The slope is \[ m=\sqrt2. \] For an ellipse, \[ y=mx\pm\sqrt{a^2m^2+b^2}. \] Substituting \[ a^2=15, \qquad b^2=6, \qquad m=\sqrt2, \] \[ y=\sqrt2\,x \pm \sqrt{15(2)+6}. \] \[ y=\sqrt2\,x \pm \sqrt{36}. \] \[ y=\sqrt2\,x\pm6. \] Hence one such tangent is \[ \boxed{y=\sqrt2\,x+6}. \]
Step 1: Use the distance between the foci. Distance between the foci is \[ 2c=6. \] Hence, \[ c=3. \] \[ c^2=9. \] \[ \cdots (1) \]
Step 2: Use the distance between the directrices. The directrices are \[ x=\pm\frac{a}{e}. \] Therefore, \[ 2\frac{a}{e}=10. \] \[ \frac{a}{e}=5. \] Since \[ e=\frac{c}{a}, \] \[ \frac{a^2}{c}=5. \] Using \[ c=3, \] \[ a^2=15. \] \[ \cdots (2) \]
Step 3: Find \(b^2\). Using \[ c^2=a^2-b^2, \] \[ 9=15-b^2. \] \[ b^2=6. \]
Step 4: Find the tangent parallel to \(y=\sqrt2\,x+5\). The slope is \[ m=\sqrt2. \] For an ellipse, \[ y=mx\pm\sqrt{a^2m^2+b^2}. \] Substituting \[ a^2=15, \qquad b^2=6, \qquad m=\sqrt2, \] \[ y=\sqrt2\,x \pm \sqrt{15(2)+6}. \] \[ y=\sqrt2\,x \pm \sqrt{36}. \] \[ y=\sqrt2\,x\pm6. \] Hence one such tangent is \[ \boxed{y=\sqrt2\,x+6}. \]
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