Question

The eccentricity at the conic section represented by the equation
\[{{\left( 10x-5 \right)}^{2}}+{{\left( 10y-5 \right)}^{2}}={{\left( 3x+4y-1 \right)}^{2}}\]
 is
A)$\dfrac{1}{2}$
B)$\dfrac{1}{\sqrt{2}}$
C)$\sqrt{2}$
D)$\dfrac{1}{10}$

Answer 1

Hint: In order to solve this question you should first simplify the equation and then find the eccentricity using the formula for eccentricity for a General curve.

Complete step by step Answer:

As discussed above we should simplify the equation of curve to get the simplified version of equation
Given,
\[{{\left( 10x-5 \right)}^{2}}+{{\left( 10y-5 \right)}^{2}}={{\left( 3x+4y-1 \right)}^{2}}\]
\[\left( 100{{x}^{2}}+25-100x \right)+\left( 100{{y}^{2}}-+25-100y \right)={{\left( 3x+4y \right)}^{2}}+1-2\left( 3x+4y \right)\]
$100{{x}^{2}}+100{{y}^{2}}-100x-100y+50=9{{x}^{2}}+16{{y}^{2}}+24xy+1-6x-8y$
$91{{x}^{2}}+84{{y}^{2}}-94x-92y-24xy+49=0$
Now rearranging this equation we get
$91{{x}^{2}}+84{{y}^{2}}-24xy-94x-92y+49=0$
Now for a given general curve
$A{{x}^{2}}+Bxy+C{{y}^{2}}+Dx+Ey+F=0$
The equation of eccentricity is given by
$e=\sqrt{\dfrac{2\sqrt{{{\left( A-C \right)}^{2}}+{{B}^{2}}}}{\left( A+C \right)+\sqrt{{{\left( A-C \right)}^{2}}+{{B}^{2}}}}}$
Now comparing the general equation to the given equation we get
\[A=91\]
\[B=-24\]
\[C=84\]
\[D=94\]
\[E=-92\]
\[F=49\]
Now substituting these values into equation of eccentricity we get
$e=\sqrt{\dfrac{2\times \sqrt{{{\left( 91-84 \right)}^{2}}+{{\left( -24 \right)}^{2}}}}{\left( 91+84 \right)+\sqrt{{{\left( 91-84 \right)}^{2}}+{{\left( -24 \right)}^{2}}}}}$
$=\sqrt{\dfrac{2\times \sqrt{{{7}^{2}}+{{24}^{2}}}}{175+\sqrt{{{7}^{2}}+{{24}^{2}}}}}$
$=\sqrt{\dfrac{2\times 25}{175+25}}$
$=\sqrt{\dfrac{50}{200}}$$=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}$
Hence the eccentricity of the curve is $\dfrac{1}{2}$
The correct option is A.
Note: In general, eccentricity means a measure of how much the deviation of the curve has occurred from the circularity of the given shape. Eccentricity for a circle is 0, for a parabola it is 1, for a hyperbola is greater than 1 and for an ellipse, it is between 0 and 1. Since in the solution the eccentricity is 0.5 which lies between 0 and 1. Therefore, the given general equation is an ellipse.
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