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Question: Coordinates of points on curve \(5{x^2} - 6xy + 5{y^2} - 4 = 0\) which are nearest to origin are \...

Coordinates of points on curve 5x2−6xy+5y2−4=05{x^2} - 6xy + 5{y^2} - 4 = 0 which are nearest to origin are
(a)(12,12)\left( a \right)\left( {\dfrac{1}{2},\dfrac{1}{2}} \right)
(b)(−12,12)\left( b \right)\left( { - \dfrac{1}{2},\dfrac{1}{2}} \right)
(c)(−12,−12)\left( c \right)\left( { - \dfrac{1}{2}, - \dfrac{1}{2}} \right)
(d)(12,−12)\left( d \right)\left( {\dfrac{1}{2}, - \dfrac{1}{2}} \right)

Explanation

Solution

Hint – In this question use the general coordinates (rcos⁡θ,rsin⁡θ)\left( {r\cos \theta ,r\sin \theta } \right) on the circumference of the given curve where r is the distance between the origin and the point (rcos⁡θ,rsin⁡θ)\left( {r\cos \theta ,r\sin \theta } \right)and θ\theta is the angle of the line joining this point and the origin from the x-axis rotate in anticlockwise or in clockwise so use these concepts to get the solution of this question.

Complete step-by-step answer:
Let the minimum distance from the origin (0, 0) is r units.
So as we know the coordinates of r is written as (rcos⁡θ,rsin⁡θ)\left( {r\cos \theta ,r\sin \theta } \right)
As the distance between origin (0, 0) and (rcos⁡θ,rsin⁡θ)\left( {r\cos \theta ,r\sin \theta } \right) is r units.
As this point is lie on the curve so it satisfies the equation of curve so substitute these points in the equation of curve we have,
x=rcos⁡θ,y=rsin⁡θx = r\cos \theta ,y = r\sin \theta
⇒5x2−6xy+5y2−4=0\Rightarrow 5{x^2} - 6xy + 5{y^2} - 4 = 0
⇒5(rcos⁡θ)2−6(rcos⁡θ)(rsin⁡θ)+5(rsin⁡θ)2−4=0\Rightarrow 5{\left( {r\cos \theta } \right)^2} - 6\left( {r\cos \theta } \right)\left( {r\sin \theta } \right) + 5{\left( {r\sin \theta } \right)^2} - 4 = 0
Now simplify this equation we have,
⇒5r2cos⁡2θ−6r2cos⁡θsin⁡θ+5r2sin⁡2θ−4=0\Rightarrow 5{r^2}{\cos ^2}\theta - 6{r^2}\cos \theta \sin \theta + 5{r^2}{\sin ^2}\theta - 4 = 0
⇒5r2(cos⁡2θ+sin⁡2θ)−3r2(2sin⁡θcos⁡θ)−4=0\Rightarrow 5{r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) - 3{r^2}\left( {2\sin \theta \cos \theta } \right) - 4 = 0
Now as we know that cos⁡2θ+sin⁡2θ{\cos ^2}\theta + {\sin ^2}\theta = 1 and 2sin⁡θcos⁡θ=sin⁡2θ2\sin \theta \cos \theta = \sin 2\theta so we have,
⇒5r2−3r2sin⁡2θ−4=0\Rightarrow 5{r^2} - 3{r^2}\sin 2\theta - 4 = 0
⇒r2(5−3sin⁡2θ)=4\Rightarrow {r^2}\left( {5 - 3\sin 2\theta } \right) = 4
⇒r2=45−3sin⁡2θ\Rightarrow {r^2} = \dfrac{4}{{5 - 3\sin 2\theta }}
Now it is given that the points are nearest to origin therefore r is minimum.
So to calculate   rmin\;{r_{min}} the value of 5−3sin⁡2θ5 - 3\sin 2\theta should be maximum.
So for (5−3sin⁡2θ)max{\left( {5 - 3\sin 2\theta } \right)_{max}} the value of sin⁡2θ\sin 2\theta should be minimum.
Now as we know that −1⩽sin⁡θ⩽1 - 1 \leqslant \sin \theta \leqslant 1.
Therefore,−1⩽sin⁡2θ⩽1 - 1 \leqslant \sin 2\theta \leqslant 1 so the minimum value of sin⁡2θ\sin 2\theta is (-1).
⇒sin⁡2θ=−1\Rightarrow \sin 2\theta = - 1
⇒2θ=sin⁡−1(−1)=sin⁡−1(sin⁡(nπ−(−1)nπ2))=nπ−(−1)nπ2\Rightarrow 2\theta = {\sin ^{ - 1}}\left( { - 1} \right) = {\sin ^{ - 1}}\left( {\sin \left( {n\pi - {{\left( { - 1} \right)}^n}\dfrac{\pi }{2}} \right)} \right) = n\pi - {\left( { - 1} \right)^n}\dfrac{\pi }{2}
Where n = 1, 2, 3, .....
⇒2θ=3π2,7π2,..............\Rightarrow 2\theta = \dfrac{{3\pi }}{2},\dfrac{{7\pi }}{2},..............
⇒θ=3π4,7π4,...............\Rightarrow \theta = \dfrac{{3\pi }}{4},\dfrac{{7\pi }}{4},...............
Therefore maximum value of (5−3sin⁡2θ)max{\left( {5 - 3\sin 2\theta } \right)_{max}}= (5 – (-3)) = (5 + 3) = 8
⇒r2=48=12\Rightarrow {r^2} = \dfrac{4}{8} = \dfrac{1}{2}
Now take square root on both sides we have,
⇒r=12\Rightarrow r = \dfrac{1}{{\sqrt 2 }}
So the coordinate are
⇒(rcos⁡θ,rsin⁡θ)=(12cos⁡3π4,12sin⁡3π4)=(−12,12)\Rightarrow \left( {r\cos \theta ,r\sin \theta } \right) = \left( {\dfrac{1}{{\sqrt 2 }}\cos \dfrac{{3\pi }}{4},\dfrac{1}{{\sqrt 2 }}\sin \dfrac{{3\pi }}{4}} \right) = \left( {\dfrac{{ - 1}}{2},\dfrac{1}{2}} \right), [∵cos⁡3π4=−12,sin⁡3π4=12]\left[ {\because \cos \dfrac{{3\pi }}{4} = - \dfrac{1}{{\sqrt 2 }},\sin \dfrac{{3\pi }}{4} = \dfrac{1}{{\sqrt 2 }}} \right]
And
⇒(rcos⁡θ,rsin⁡θ)=(12cos⁡7π4,12sin⁡7π4)=(12,−12)\Rightarrow \left( {r\cos \theta ,r\sin \theta } \right) = \left( {\dfrac{1}{{\sqrt 2 }}\cos \dfrac{{7\pi }}{4},\dfrac{1}{{\sqrt 2 }}\sin \dfrac{{7\pi }}{4}} \right) = \left( {\dfrac{1}{2},\dfrac{{ - 1}}{2}} \right), [∵cos⁡7π4=12,sin⁡7π4=−12]\left[ {\because \cos \dfrac{{7\pi }}{4} = \dfrac{1}{{\sqrt 2 }},\sin \dfrac{{7\pi }}{4} = - \dfrac{1}{{\sqrt 2 }}} \right]
So these are the required coordinates which are nearest to the origin.
Hence options (B) and (D) are correct.

Note – Whenever we face such types of questions the key concept is the general parametric coordinates on the circumference of the curve which is written above then satisfy this coordinates on the curve and calculate the value of r in terms of θ\theta then we have to calculate the minimum value of r so we have to maximize the denominator as above then calculate the value of r and θ\theta as above and substitute these values in the (rcos⁡θ,rsin⁡θ)\left( {r\cos \theta ,r\sin \theta } \right) we will get the required coordinates.