Question:
The length of the tangent drawn at the point \(P(1, 3\sqrt{3})\) on the curve is \(\frac{x^2}{3} + \frac{y^2}{27} = 4\):
The length of the tangent drawn at the point \(P(1, 3\sqrt{3})\) on the curve is \(\frac{x^2}{3} + \frac{y^2}{27} = 4\):
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To find the length of a tangent segment intercepted by an axis, find the equation of the tangent line first, calculate the intercept coordinate, and then apply the distance formula between the contact point and the intercept point.
Updated On: Jun 9, 2026
- \( 4 \)
- \( 6 \)
- \( 12 \)
- \( 8 \)
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The Correct Option is B
Solution and Explanation
Concept:
For an ellipse defined by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), the slope of the tangent at any point \((x_1, y_1)\) is given by differentiating the equation with respect to \(x\). The length of the tangent segment intercepted by the axes is calculated using the distance formula between the point of tangency and the x-intercept.
Step 1: Verify the point \(P\) and find the derivative \(\frac{dy}{dx}\).
The given equation is \(\frac{x^2}{3} + \frac{y^2}{27} = 4\). Differentiating with respect to \(x\): \[ \frac{2x}{3} + \frac{2y}{27} \frac{dy}{dx} = 0 \] \[ \frac{dy}{dx} = -\frac{2x}{3} \cdot \frac{27}{2y} = -\frac{9x}{y} \] At point \(P(1, 3\sqrt{3})\): \[ \left. \frac{dy}{dx} \right|_{(1, 3\sqrt{3})} = -\frac{9(1)}{3\sqrt{3}} = -\frac{3}{\sqrt{3}} = -\sqrt{3} \]
Step 2: Find the equation of the tangent line at \(P\).
Using the point-slope form \(y - y_1 = m(x - x_1)\): \[ y - 3\sqrt{3} = -\sqrt{3}(x - 1) \] \[ y - 3\sqrt{3} = -\sqrt{3}x + \sqrt{3} \] \[ \sqrt{3}x + y = 4\sqrt{3} \]
Step 3: Find the x-intercept and calculate the distance to \(P\).
To find the x-intercept, set \(y = 0\): \[ \sqrt{3}x = 4\sqrt{3} \implies x = 4 \] So, the tangent meets the x-axis at \(A(4, 0)\). The length of the tangent segment \(PA\) is: \[ PA = \sqrt{(4 - 1)^2 + (0 - 3\sqrt{3})^2} \] \[ PA = \sqrt{3^2 + (-3\sqrt{3})^2} \] \[ PA = \sqrt{9 + (9 \times 3)} = \sqrt{9 + 27} = \sqrt{36} = 6 \]
Conclusion: The length of the tangent segment intercepted by the x-axis is 6 units. center Final Answer: (B) 6 center
Step 1: Verify the point \(P\) and find the derivative \(\frac{dy}{dx}\).
The given equation is \(\frac{x^2}{3} + \frac{y^2}{27} = 4\). Differentiating with respect to \(x\): \[ \frac{2x}{3} + \frac{2y}{27} \frac{dy}{dx} = 0 \] \[ \frac{dy}{dx} = -\frac{2x}{3} \cdot \frac{27}{2y} = -\frac{9x}{y} \] At point \(P(1, 3\sqrt{3})\): \[ \left. \frac{dy}{dx} \right|_{(1, 3\sqrt{3})} = -\frac{9(1)}{3\sqrt{3}} = -\frac{3}{\sqrt{3}} = -\sqrt{3} \]
Step 2: Find the equation of the tangent line at \(P\).
Using the point-slope form \(y - y_1 = m(x - x_1)\): \[ y - 3\sqrt{3} = -\sqrt{3}(x - 1) \] \[ y - 3\sqrt{3} = -\sqrt{3}x + \sqrt{3} \] \[ \sqrt{3}x + y = 4\sqrt{3} \]
Step 3: Find the x-intercept and calculate the distance to \(P\).
To find the x-intercept, set \(y = 0\): \[ \sqrt{3}x = 4\sqrt{3} \implies x = 4 \] So, the tangent meets the x-axis at \(A(4, 0)\). The length of the tangent segment \(PA\) is: \[ PA = \sqrt{(4 - 1)^2 + (0 - 3\sqrt{3})^2} \] \[ PA = \sqrt{3^2 + (-3\sqrt{3})^2} \] \[ PA = \sqrt{9 + (9 \times 3)} = \sqrt{9 + 27} = \sqrt{36} = 6 \]
Conclusion: The length of the tangent segment intercepted by the x-axis is 6 units. center Final Answer: (B) 6 center
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