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An equilateral triangle is inscribed in ...

An equilateral triangle is inscribed in the ellipse `x^2+3y^2=3` such that one vertex of the triangle is (0, 1) and one altitude of the triangle is along the y-axis. The length of its side is

A

`4sqrt3//5`

B

`3sqrt3//5`

C

`6sqrt3//5`

D

`2sqrt3`

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The correct Answer is:
To find the length of the side of the equilateral triangle inscribed in the ellipse given by the equation \(x^2 + 3y^2 = 3\), we will follow these steps: ### Step 1: Rewrite the equation of the ellipse The equation of the ellipse can be rewritten in standard form: \[ \frac{x^2}{3} + \frac{y^2}{1} = 1 \] This indicates that the semi-major axis \(a = \sqrt{3}\) and the semi-minor axis \(b = 1\). ### Step 2: Identify the vertex and altitude We know one vertex of the triangle is at \(C(0, 1)\) and that one altitude is along the y-axis. This means the other two vertices \(A\) and \(B\) will be symmetric about the y-axis. ### Step 3: Determine the coordinates of points A and B Let the coordinates of points \(A\) and \(B\) be \((-x, y_A)\) and \((x, y_A)\) respectively. Since the triangle is equilateral, the altitude from \(C\) to the base \(AB\) will bisect \(AB\) at point \(D(0, y_A)\). ### Step 4: Find the relationship between the coordinates The length of the altitude \(CD\) can be calculated as: \[ CD = 1 - y_A \] The length of the side \(AB\) can be calculated using the distance formula: \[ AB = \sqrt{((-x) - x)^2 + (y_A - y_A)^2} = 2x \] ### Step 5: Use the properties of the equilateral triangle In an equilateral triangle, the relationship between the side length \(s\) and the altitude \(h\) is given by: \[ h = \frac{\sqrt{3}}{2}s \] Substituting \(s = 2x\) into the equation gives: \[ 1 - y_A = \frac{\sqrt{3}}{2}(2x) \] Thus, we have: \[ 1 - y_A = \sqrt{3}x \] From this, we can express \(y_A\) as: \[ y_A = 1 - \sqrt{3}x \] ### Step 6: Substitute into the ellipse equation Substituting \(y_A\) into the ellipse equation: \[ x^2 + 3(1 - \sqrt{3}x)^2 = 3 \] Expanding this: \[ x^2 + 3(1 - 2\sqrt{3}x + 3x^2) = 3 \] \[ x^2 + 3 - 6\sqrt{3}x + 9x^2 = 3 \] Combining like terms: \[ 10x^2 - 6\sqrt{3}x + 0 = 0 \] ### Step 7: Solve the quadratic equation Factoring out \(2x\): \[ 2x(5x - 3\sqrt{3}) = 0 \] This gives us \(x = 0\) or \(x = \frac{3\sqrt{3}}{5}\). ### Step 8: Find the coordinates of A and B Using \(x = \frac{3\sqrt{3}}{5}\): \[ y_A = 1 - \sqrt{3}\left(\frac{3\sqrt{3}}{5}\right) = 1 - \frac{9}{5} = -\frac{4}{5} \] Thus, the coordinates of \(A\) and \(B\) are: \[ A\left(-\frac{3\sqrt{3}}{5}, -\frac{4}{5}\right), \quad B\left(\frac{3\sqrt{3}}{5}, -\frac{4}{5}\right) \] ### Step 9: Calculate the side length The length of the side \(s\) is: \[ s = 2x = 2\left(\frac{3\sqrt{3}}{5}\right) = \frac{6\sqrt{3}}{5} \] ### Final Answer The length of the side of the equilateral triangle is: \[ \frac{6\sqrt{3}}{5} \]
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MCGROW HILL PUBLICATION-ELLIPSE-Exercise (Level 1 Single Correct)
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  3. An equilateral triangle is inscribed in the ellipse x^2+3y^2=3 such t...

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  6. The eccentricity of an ellipse, with centre at the origin is 2/3. If o...

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  7. The locus of the foot of the perpendicular from the foci an any tangen...

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  8. If the ellipse x^2/a^2+y^2/b^2=1 and the circle x^2+y^2=r^2 where bltr...

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  9. The product of the perpendiculars from the foci of the ellipse x^2/14...

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  13. If one extremity of the minor axis of the ellipse x^2/a^2+y^2/b^2=1 a...

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  14. The length of the major axis of the ellipse (5x-10)^2 +(5y+13)^2 = (3x...

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  16. Let P be any point on a directrix of an ellipse of eccentricity e, S b...

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  17. If a tangent of slope 2 of the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 is no...

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