87074
A straight line \(L_{1}\) passing through \(A(3,1)\) meets the coordinate axes at \(P\) and \(Q\) such that its distance from the origin \(O\) is maximum. Then area of \(\triangle \mathrm{OPQ}\) is sq. units
1 \(\frac{100}{3}\)
2 \(\frac{25}{3}\)
3 \(\frac{50}{3}\)
4 \(\frac{200}{3}\)
Explanation:
(C) : : Let, Line \(A B\) will be farthest from the origin if op is right angled to line drawn \(\mathrm{M}_{\mathrm{OP}}=\frac{1}{3}\) \(\mathrm{M}_{\mathrm{AP}}=-3\) Thus, the equation of \(A B\) is \((y-1)=-3(x-3)\) \(\mathrm{A}=\left(\frac{10}{3}, 0\right), \mathrm{B}=(0,10)\) \(\Delta \mathrm{OAB}=\frac{1}{2}(\mathrm{OA}),(\mathrm{OB})=\frac{1}{2} \times \frac{10}{3} \times 10=\frac{100}{6}=\frac{50}{3}\)
AP EAMCET-2020-18.09.2020
Application of the Integrals
87075
Let \(A\) be the area of in-circle and \(A_{1}, A_{2}, A_{3}\) be the area of ex-circles of a triangle. If \(A_{1}=4, A_{2}\) \(=9, A_{3}=16\), then \(A=\)
87076
The area of the greatest rectangle that can be inscribed in an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is (in square units)
1 \(\pi \mathrm{ab}\)
2 \(\mathrm{ab}\)
3 \(2 \mathrm{ab}\)
4 \(4 \mathrm{ab}\)
Explanation:
(C) : : As, we know that, Equation of ellipse \(\frac{\mathrm{x}^{2}}{1^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1\) Let, the four vertices of rectangle lie on ellipse is \(\mathrm{P}(\mathrm{a}\) \(\cos \theta,-\mathrm{b} \sin \theta), \theta(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta), \mathrm{R}(-\mathrm{a} \cos \theta, \mathrm{b} \sin\) \(\theta)\) and \(S(-a \cos \theta,-b \sin \theta)\) Area of rectangle \(\mathrm{PQRS}=\mathrm{PQ} \times \mathrm{RS}\) \(A=4 a b \sin \theta \cos \theta\) \(\mathrm{A}=2 \mathrm{ab} \sin 2 \theta\) \(\frac{\mathrm{dA}}{\mathrm{d} \theta}=4 \mathrm{ab} \cos 2 \theta\) For maxima or minima put \(\frac{\mathrm{d} A}{\mathrm{~d} \theta}=0\) \(4 \mathrm{ab} \cos 2 \theta=0\) \(\cos 2 \theta=0\) \(2 \theta=\frac{\pi}{2}\) \(\frac{d^{2} A}{d \theta^{2}}=-8 a b \sin 2 \theta\lt 0\) Therefore, Area is maximum of \(2 \theta=\frac{\pi}{2}\) \(\therefore \mathrm{A}=2 \mathrm{ab} \sin \frac{\pi}{2}\) \(\mathrm{A}=2 \mathrm{ab}\) \(\left[\because \sin \frac{\pi}{2}=1\right]\)
[JCECE-2018]
Application of the Integrals
87039
The area (in sq units) between the curve \(y^{2}=\) \(8 x\) and its latus rectum is
87074
A straight line \(L_{1}\) passing through \(A(3,1)\) meets the coordinate axes at \(P\) and \(Q\) such that its distance from the origin \(O\) is maximum. Then area of \(\triangle \mathrm{OPQ}\) is sq. units
1 \(\frac{100}{3}\)
2 \(\frac{25}{3}\)
3 \(\frac{50}{3}\)
4 \(\frac{200}{3}\)
Explanation:
(C) : : Let, Line \(A B\) will be farthest from the origin if op is right angled to line drawn \(\mathrm{M}_{\mathrm{OP}}=\frac{1}{3}\) \(\mathrm{M}_{\mathrm{AP}}=-3\) Thus, the equation of \(A B\) is \((y-1)=-3(x-3)\) \(\mathrm{A}=\left(\frac{10}{3}, 0\right), \mathrm{B}=(0,10)\) \(\Delta \mathrm{OAB}=\frac{1}{2}(\mathrm{OA}),(\mathrm{OB})=\frac{1}{2} \times \frac{10}{3} \times 10=\frac{100}{6}=\frac{50}{3}\)
AP EAMCET-2020-18.09.2020
Application of the Integrals
87075
Let \(A\) be the area of in-circle and \(A_{1}, A_{2}, A_{3}\) be the area of ex-circles of a triangle. If \(A_{1}=4, A_{2}\) \(=9, A_{3}=16\), then \(A=\)
87076
The area of the greatest rectangle that can be inscribed in an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is (in square units)
1 \(\pi \mathrm{ab}\)
2 \(\mathrm{ab}\)
3 \(2 \mathrm{ab}\)
4 \(4 \mathrm{ab}\)
Explanation:
(C) : : As, we know that, Equation of ellipse \(\frac{\mathrm{x}^{2}}{1^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1\) Let, the four vertices of rectangle lie on ellipse is \(\mathrm{P}(\mathrm{a}\) \(\cos \theta,-\mathrm{b} \sin \theta), \theta(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta), \mathrm{R}(-\mathrm{a} \cos \theta, \mathrm{b} \sin\) \(\theta)\) and \(S(-a \cos \theta,-b \sin \theta)\) Area of rectangle \(\mathrm{PQRS}=\mathrm{PQ} \times \mathrm{RS}\) \(A=4 a b \sin \theta \cos \theta\) \(\mathrm{A}=2 \mathrm{ab} \sin 2 \theta\) \(\frac{\mathrm{dA}}{\mathrm{d} \theta}=4 \mathrm{ab} \cos 2 \theta\) For maxima or minima put \(\frac{\mathrm{d} A}{\mathrm{~d} \theta}=0\) \(4 \mathrm{ab} \cos 2 \theta=0\) \(\cos 2 \theta=0\) \(2 \theta=\frac{\pi}{2}\) \(\frac{d^{2} A}{d \theta^{2}}=-8 a b \sin 2 \theta\lt 0\) Therefore, Area is maximum of \(2 \theta=\frac{\pi}{2}\) \(\therefore \mathrm{A}=2 \mathrm{ab} \sin \frac{\pi}{2}\) \(\mathrm{A}=2 \mathrm{ab}\) \(\left[\because \sin \frac{\pi}{2}=1\right]\)
[JCECE-2018]
Application of the Integrals
87039
The area (in sq units) between the curve \(y^{2}=\) \(8 x\) and its latus rectum is
87074
A straight line \(L_{1}\) passing through \(A(3,1)\) meets the coordinate axes at \(P\) and \(Q\) such that its distance from the origin \(O\) is maximum. Then area of \(\triangle \mathrm{OPQ}\) is sq. units
1 \(\frac{100}{3}\)
2 \(\frac{25}{3}\)
3 \(\frac{50}{3}\)
4 \(\frac{200}{3}\)
Explanation:
(C) : : Let, Line \(A B\) will be farthest from the origin if op is right angled to line drawn \(\mathrm{M}_{\mathrm{OP}}=\frac{1}{3}\) \(\mathrm{M}_{\mathrm{AP}}=-3\) Thus, the equation of \(A B\) is \((y-1)=-3(x-3)\) \(\mathrm{A}=\left(\frac{10}{3}, 0\right), \mathrm{B}=(0,10)\) \(\Delta \mathrm{OAB}=\frac{1}{2}(\mathrm{OA}),(\mathrm{OB})=\frac{1}{2} \times \frac{10}{3} \times 10=\frac{100}{6}=\frac{50}{3}\)
AP EAMCET-2020-18.09.2020
Application of the Integrals
87075
Let \(A\) be the area of in-circle and \(A_{1}, A_{2}, A_{3}\) be the area of ex-circles of a triangle. If \(A_{1}=4, A_{2}\) \(=9, A_{3}=16\), then \(A=\)
87076
The area of the greatest rectangle that can be inscribed in an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is (in square units)
1 \(\pi \mathrm{ab}\)
2 \(\mathrm{ab}\)
3 \(2 \mathrm{ab}\)
4 \(4 \mathrm{ab}\)
Explanation:
(C) : : As, we know that, Equation of ellipse \(\frac{\mathrm{x}^{2}}{1^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1\) Let, the four vertices of rectangle lie on ellipse is \(\mathrm{P}(\mathrm{a}\) \(\cos \theta,-\mathrm{b} \sin \theta), \theta(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta), \mathrm{R}(-\mathrm{a} \cos \theta, \mathrm{b} \sin\) \(\theta)\) and \(S(-a \cos \theta,-b \sin \theta)\) Area of rectangle \(\mathrm{PQRS}=\mathrm{PQ} \times \mathrm{RS}\) \(A=4 a b \sin \theta \cos \theta\) \(\mathrm{A}=2 \mathrm{ab} \sin 2 \theta\) \(\frac{\mathrm{dA}}{\mathrm{d} \theta}=4 \mathrm{ab} \cos 2 \theta\) For maxima or minima put \(\frac{\mathrm{d} A}{\mathrm{~d} \theta}=0\) \(4 \mathrm{ab} \cos 2 \theta=0\) \(\cos 2 \theta=0\) \(2 \theta=\frac{\pi}{2}\) \(\frac{d^{2} A}{d \theta^{2}}=-8 a b \sin 2 \theta\lt 0\) Therefore, Area is maximum of \(2 \theta=\frac{\pi}{2}\) \(\therefore \mathrm{A}=2 \mathrm{ab} \sin \frac{\pi}{2}\) \(\mathrm{A}=2 \mathrm{ab}\) \(\left[\because \sin \frac{\pi}{2}=1\right]\)
[JCECE-2018]
Application of the Integrals
87039
The area (in sq units) between the curve \(y^{2}=\) \(8 x\) and its latus rectum is
87074
A straight line \(L_{1}\) passing through \(A(3,1)\) meets the coordinate axes at \(P\) and \(Q\) such that its distance from the origin \(O\) is maximum. Then area of \(\triangle \mathrm{OPQ}\) is sq. units
1 \(\frac{100}{3}\)
2 \(\frac{25}{3}\)
3 \(\frac{50}{3}\)
4 \(\frac{200}{3}\)
Explanation:
(C) : : Let, Line \(A B\) will be farthest from the origin if op is right angled to line drawn \(\mathrm{M}_{\mathrm{OP}}=\frac{1}{3}\) \(\mathrm{M}_{\mathrm{AP}}=-3\) Thus, the equation of \(A B\) is \((y-1)=-3(x-3)\) \(\mathrm{A}=\left(\frac{10}{3}, 0\right), \mathrm{B}=(0,10)\) \(\Delta \mathrm{OAB}=\frac{1}{2}(\mathrm{OA}),(\mathrm{OB})=\frac{1}{2} \times \frac{10}{3} \times 10=\frac{100}{6}=\frac{50}{3}\)
AP EAMCET-2020-18.09.2020
Application of the Integrals
87075
Let \(A\) be the area of in-circle and \(A_{1}, A_{2}, A_{3}\) be the area of ex-circles of a triangle. If \(A_{1}=4, A_{2}\) \(=9, A_{3}=16\), then \(A=\)
87076
The area of the greatest rectangle that can be inscribed in an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is (in square units)
1 \(\pi \mathrm{ab}\)
2 \(\mathrm{ab}\)
3 \(2 \mathrm{ab}\)
4 \(4 \mathrm{ab}\)
Explanation:
(C) : : As, we know that, Equation of ellipse \(\frac{\mathrm{x}^{2}}{1^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1\) Let, the four vertices of rectangle lie on ellipse is \(\mathrm{P}(\mathrm{a}\) \(\cos \theta,-\mathrm{b} \sin \theta), \theta(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta), \mathrm{R}(-\mathrm{a} \cos \theta, \mathrm{b} \sin\) \(\theta)\) and \(S(-a \cos \theta,-b \sin \theta)\) Area of rectangle \(\mathrm{PQRS}=\mathrm{PQ} \times \mathrm{RS}\) \(A=4 a b \sin \theta \cos \theta\) \(\mathrm{A}=2 \mathrm{ab} \sin 2 \theta\) \(\frac{\mathrm{dA}}{\mathrm{d} \theta}=4 \mathrm{ab} \cos 2 \theta\) For maxima or minima put \(\frac{\mathrm{d} A}{\mathrm{~d} \theta}=0\) \(4 \mathrm{ab} \cos 2 \theta=0\) \(\cos 2 \theta=0\) \(2 \theta=\frac{\pi}{2}\) \(\frac{d^{2} A}{d \theta^{2}}=-8 a b \sin 2 \theta\lt 0\) Therefore, Area is maximum of \(2 \theta=\frac{\pi}{2}\) \(\therefore \mathrm{A}=2 \mathrm{ab} \sin \frac{\pi}{2}\) \(\mathrm{A}=2 \mathrm{ab}\) \(\left[\because \sin \frac{\pi}{2}=1\right]\)
[JCECE-2018]
Application of the Integrals
87039
The area (in sq units) between the curve \(y^{2}=\) \(8 x\) and its latus rectum is
