(B) : Given, Equation of tangent drawn to the ellipse \(2 x^{2}+3 y^{2}=5\) \(\therefore \quad \mathrm{y}=\mathrm{mx} \pm \sqrt{\frac{5}{2} \mathrm{~m}^{2}+\frac{5}{3}}\) It is passes through \((1,3)\) \(3=m \pm \sqrt{\frac{5}{2} m^{2}+\frac{5}{3}}\) \(3 m^{2}+12 m-\frac{44}{3}=0\) Let \(\theta\) be the angle between the tangent \(\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \Rightarrow \tan \theta=\left|\frac{3 \sqrt{320}}{-35}\right|\) \(\theta=\tan ^{-1}\left(\frac{24}{7 \sqrt{5}}\right)\)
JEE Main-2022-26.07.2022
Application of Derivatives
85788
The acute angle between the tangents drawn at the point of intersection (other than the origin) of the curves \(x^{2}=4 y\) and \(y^{2}=4 x\) is
1 \(\tan ^{-1}\left(\frac{1}{2}\right)\)
2 \(\sin ^{-1}\left(\frac{3}{5}\right)\)
3 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
4 \(\tan ^{-1}\left(\frac{2}{3}\right)\)
Explanation:
(B) : Given that, \(y^{2}=4 x\) Differentiating w.r.t. \(\mathrm{x}\), we get - \(2 y \frac{d y}{d x}=4\) \(\frac{d y}{d x}=\frac{2}{y} \tag{i}\) And, \(\quad x^{2}=4 y\) Differentiating w.r.t. \(\mathrm{x}\), we get - \(2 x=4 \frac{d y}{d x}\) \(\frac{d y}{d x}=\frac{x}{2} \tag{ii}\) Then, the slope of tangent at \((0,0)\) of equation (i) is parallel to y axis. And, the slope of tangent at \((0,0)\) of equation (ii) parallel to \(\mathrm{x}\) - axis. But, given the point of intersection (other than the origin) Then, let the slope of tangent at \((1,1)\) of equation (i)- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}_{1}=\frac{2}{1}=2\) And, the slope of tangent at \((1,1)\) of equation (ii) is, \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}_{2}=\frac{1}{2}\) Then, \(\theta\) be the angle between the tangent \(\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \Rightarrow \tan \theta=\left|\frac{2-\frac{1}{2}}{1+2 \times \frac{1}{2}}\right|\) \(\tan \theta=\left|\frac{\frac{3}{2}}{1+1}\right| \Rightarrow \tan =\frac{3}{4}\) Then, \(\quad \sin \theta=\frac{3}{5}\) So, \(\theta=\sin ^{-1}\left(\frac{3}{5}\right)\)
AP EAMCET-2018-22.04.2018
Application of Derivatives
85789
Two sides of a triangle are given. If the area of the triangle is maximum then the angle between the given sides is
1 \(45^{\circ}\)
2 \(30^{\circ}\)
3 \(60^{\circ}\)
4 \(90^{\circ}\)
Explanation:
(D) : Let, \(x\) and \(y\) be the length of given sides and \(\theta\) be angle between them. Let, \(A\) be the area of the triangle. Then, \(\quad A=\frac{1}{2} x y \sin \theta \Rightarrow A(\theta)=\frac{1}{2} x y \sin \theta\) \(A^{\prime}(\theta)=\frac{1}{2} x y \cos \theta \Rightarrow A^{\prime \prime}(\theta)=-\frac{1}{2} x y \sin \theta\) For maxima \(/\) minima \(\mathrm{A}^{\prime}(\theta)=0\) \(\frac{1}{2}(x y \cos \theta)=0\) \(\cos \theta=0\) \(\theta=\frac{\pi}{2}\) \(A^{\prime \prime}\left(\frac{\pi}{2}\right)=-\frac{1}{2} x y \sin \frac{\pi}{2}=-\frac{1}{2} x y\lt 0\) So, area is maximum at \(\theta=\frac{\pi}{2}\) or \(90^{\circ}\)
AP EAMCET-2022-06.07.2022
Application of Derivatives
85790
The angle between the tangents drawn at \((0,0)\) to the curves \(y^{3}-x^{2} y+5 y-2 x=0\) and \(x^{4}-x^{3} y^{2}+5 x+2 y=0\) is
1 \(\pi / 6\)
2 \(\pi / 4\)
3 \(\pi / 3\)
4 \(\pi / 2\)
Explanation:
(D) : Given that, \(y^{3}-x^{2} y+5 y-2 x=0\) Slope of tangent to curve at \((10,0)\) \(\mathrm{m}_{1}=\frac{2}{5}\) For second curve is \(x^{4}-x^{3} y^{2}+5 x+2 y=0\) Slope of tangent to curve at \((0,0)\) \(\mathrm{m}_{2}=-\frac{5}{2}\) Therefore, \(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=\frac{2}{5} \times \frac{-5}{2}=-1\) So, angle between tangent to the curves is \(\frac{\pi}{2}\).
(B) : Given, Equation of tangent drawn to the ellipse \(2 x^{2}+3 y^{2}=5\) \(\therefore \quad \mathrm{y}=\mathrm{mx} \pm \sqrt{\frac{5}{2} \mathrm{~m}^{2}+\frac{5}{3}}\) It is passes through \((1,3)\) \(3=m \pm \sqrt{\frac{5}{2} m^{2}+\frac{5}{3}}\) \(3 m^{2}+12 m-\frac{44}{3}=0\) Let \(\theta\) be the angle between the tangent \(\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \Rightarrow \tan \theta=\left|\frac{3 \sqrt{320}}{-35}\right|\) \(\theta=\tan ^{-1}\left(\frac{24}{7 \sqrt{5}}\right)\)
JEE Main-2022-26.07.2022
Application of Derivatives
85788
The acute angle between the tangents drawn at the point of intersection (other than the origin) of the curves \(x^{2}=4 y\) and \(y^{2}=4 x\) is
1 \(\tan ^{-1}\left(\frac{1}{2}\right)\)
2 \(\sin ^{-1}\left(\frac{3}{5}\right)\)
3 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
4 \(\tan ^{-1}\left(\frac{2}{3}\right)\)
Explanation:
(B) : Given that, \(y^{2}=4 x\) Differentiating w.r.t. \(\mathrm{x}\), we get - \(2 y \frac{d y}{d x}=4\) \(\frac{d y}{d x}=\frac{2}{y} \tag{i}\) And, \(\quad x^{2}=4 y\) Differentiating w.r.t. \(\mathrm{x}\), we get - \(2 x=4 \frac{d y}{d x}\) \(\frac{d y}{d x}=\frac{x}{2} \tag{ii}\) Then, the slope of tangent at \((0,0)\) of equation (i) is parallel to y axis. And, the slope of tangent at \((0,0)\) of equation (ii) parallel to \(\mathrm{x}\) - axis. But, given the point of intersection (other than the origin) Then, let the slope of tangent at \((1,1)\) of equation (i)- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}_{1}=\frac{2}{1}=2\) And, the slope of tangent at \((1,1)\) of equation (ii) is, \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}_{2}=\frac{1}{2}\) Then, \(\theta\) be the angle between the tangent \(\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \Rightarrow \tan \theta=\left|\frac{2-\frac{1}{2}}{1+2 \times \frac{1}{2}}\right|\) \(\tan \theta=\left|\frac{\frac{3}{2}}{1+1}\right| \Rightarrow \tan =\frac{3}{4}\) Then, \(\quad \sin \theta=\frac{3}{5}\) So, \(\theta=\sin ^{-1}\left(\frac{3}{5}\right)\)
AP EAMCET-2018-22.04.2018
Application of Derivatives
85789
Two sides of a triangle are given. If the area of the triangle is maximum then the angle between the given sides is
1 \(45^{\circ}\)
2 \(30^{\circ}\)
3 \(60^{\circ}\)
4 \(90^{\circ}\)
Explanation:
(D) : Let, \(x\) and \(y\) be the length of given sides and \(\theta\) be angle between them. Let, \(A\) be the area of the triangle. Then, \(\quad A=\frac{1}{2} x y \sin \theta \Rightarrow A(\theta)=\frac{1}{2} x y \sin \theta\) \(A^{\prime}(\theta)=\frac{1}{2} x y \cos \theta \Rightarrow A^{\prime \prime}(\theta)=-\frac{1}{2} x y \sin \theta\) For maxima \(/\) minima \(\mathrm{A}^{\prime}(\theta)=0\) \(\frac{1}{2}(x y \cos \theta)=0\) \(\cos \theta=0\) \(\theta=\frac{\pi}{2}\) \(A^{\prime \prime}\left(\frac{\pi}{2}\right)=-\frac{1}{2} x y \sin \frac{\pi}{2}=-\frac{1}{2} x y\lt 0\) So, area is maximum at \(\theta=\frac{\pi}{2}\) or \(90^{\circ}\)
AP EAMCET-2022-06.07.2022
Application of Derivatives
85790
The angle between the tangents drawn at \((0,0)\) to the curves \(y^{3}-x^{2} y+5 y-2 x=0\) and \(x^{4}-x^{3} y^{2}+5 x+2 y=0\) is
1 \(\pi / 6\)
2 \(\pi / 4\)
3 \(\pi / 3\)
4 \(\pi / 2\)
Explanation:
(D) : Given that, \(y^{3}-x^{2} y+5 y-2 x=0\) Slope of tangent to curve at \((10,0)\) \(\mathrm{m}_{1}=\frac{2}{5}\) For second curve is \(x^{4}-x^{3} y^{2}+5 x+2 y=0\) Slope of tangent to curve at \((0,0)\) \(\mathrm{m}_{2}=-\frac{5}{2}\) Therefore, \(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=\frac{2}{5} \times \frac{-5}{2}=-1\) So, angle between tangent to the curves is \(\frac{\pi}{2}\).
(B) : Given, Equation of tangent drawn to the ellipse \(2 x^{2}+3 y^{2}=5\) \(\therefore \quad \mathrm{y}=\mathrm{mx} \pm \sqrt{\frac{5}{2} \mathrm{~m}^{2}+\frac{5}{3}}\) It is passes through \((1,3)\) \(3=m \pm \sqrt{\frac{5}{2} m^{2}+\frac{5}{3}}\) \(3 m^{2}+12 m-\frac{44}{3}=0\) Let \(\theta\) be the angle between the tangent \(\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \Rightarrow \tan \theta=\left|\frac{3 \sqrt{320}}{-35}\right|\) \(\theta=\tan ^{-1}\left(\frac{24}{7 \sqrt{5}}\right)\)
JEE Main-2022-26.07.2022
Application of Derivatives
85788
The acute angle between the tangents drawn at the point of intersection (other than the origin) of the curves \(x^{2}=4 y\) and \(y^{2}=4 x\) is
1 \(\tan ^{-1}\left(\frac{1}{2}\right)\)
2 \(\sin ^{-1}\left(\frac{3}{5}\right)\)
3 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
4 \(\tan ^{-1}\left(\frac{2}{3}\right)\)
Explanation:
(B) : Given that, \(y^{2}=4 x\) Differentiating w.r.t. \(\mathrm{x}\), we get - \(2 y \frac{d y}{d x}=4\) \(\frac{d y}{d x}=\frac{2}{y} \tag{i}\) And, \(\quad x^{2}=4 y\) Differentiating w.r.t. \(\mathrm{x}\), we get - \(2 x=4 \frac{d y}{d x}\) \(\frac{d y}{d x}=\frac{x}{2} \tag{ii}\) Then, the slope of tangent at \((0,0)\) of equation (i) is parallel to y axis. And, the slope of tangent at \((0,0)\) of equation (ii) parallel to \(\mathrm{x}\) - axis. But, given the point of intersection (other than the origin) Then, let the slope of tangent at \((1,1)\) of equation (i)- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}_{1}=\frac{2}{1}=2\) And, the slope of tangent at \((1,1)\) of equation (ii) is, \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}_{2}=\frac{1}{2}\) Then, \(\theta\) be the angle between the tangent \(\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \Rightarrow \tan \theta=\left|\frac{2-\frac{1}{2}}{1+2 \times \frac{1}{2}}\right|\) \(\tan \theta=\left|\frac{\frac{3}{2}}{1+1}\right| \Rightarrow \tan =\frac{3}{4}\) Then, \(\quad \sin \theta=\frac{3}{5}\) So, \(\theta=\sin ^{-1}\left(\frac{3}{5}\right)\)
AP EAMCET-2018-22.04.2018
Application of Derivatives
85789
Two sides of a triangle are given. If the area of the triangle is maximum then the angle between the given sides is
1 \(45^{\circ}\)
2 \(30^{\circ}\)
3 \(60^{\circ}\)
4 \(90^{\circ}\)
Explanation:
(D) : Let, \(x\) and \(y\) be the length of given sides and \(\theta\) be angle between them. Let, \(A\) be the area of the triangle. Then, \(\quad A=\frac{1}{2} x y \sin \theta \Rightarrow A(\theta)=\frac{1}{2} x y \sin \theta\) \(A^{\prime}(\theta)=\frac{1}{2} x y \cos \theta \Rightarrow A^{\prime \prime}(\theta)=-\frac{1}{2} x y \sin \theta\) For maxima \(/\) minima \(\mathrm{A}^{\prime}(\theta)=0\) \(\frac{1}{2}(x y \cos \theta)=0\) \(\cos \theta=0\) \(\theta=\frac{\pi}{2}\) \(A^{\prime \prime}\left(\frac{\pi}{2}\right)=-\frac{1}{2} x y \sin \frac{\pi}{2}=-\frac{1}{2} x y\lt 0\) So, area is maximum at \(\theta=\frac{\pi}{2}\) or \(90^{\circ}\)
AP EAMCET-2022-06.07.2022
Application of Derivatives
85790
The angle between the tangents drawn at \((0,0)\) to the curves \(y^{3}-x^{2} y+5 y-2 x=0\) and \(x^{4}-x^{3} y^{2}+5 x+2 y=0\) is
1 \(\pi / 6\)
2 \(\pi / 4\)
3 \(\pi / 3\)
4 \(\pi / 2\)
Explanation:
(D) : Given that, \(y^{3}-x^{2} y+5 y-2 x=0\) Slope of tangent to curve at \((10,0)\) \(\mathrm{m}_{1}=\frac{2}{5}\) For second curve is \(x^{4}-x^{3} y^{2}+5 x+2 y=0\) Slope of tangent to curve at \((0,0)\) \(\mathrm{m}_{2}=-\frac{5}{2}\) Therefore, \(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=\frac{2}{5} \times \frac{-5}{2}=-1\) So, angle between tangent to the curves is \(\frac{\pi}{2}\).
(B) : Given, Equation of tangent drawn to the ellipse \(2 x^{2}+3 y^{2}=5\) \(\therefore \quad \mathrm{y}=\mathrm{mx} \pm \sqrt{\frac{5}{2} \mathrm{~m}^{2}+\frac{5}{3}}\) It is passes through \((1,3)\) \(3=m \pm \sqrt{\frac{5}{2} m^{2}+\frac{5}{3}}\) \(3 m^{2}+12 m-\frac{44}{3}=0\) Let \(\theta\) be the angle between the tangent \(\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \Rightarrow \tan \theta=\left|\frac{3 \sqrt{320}}{-35}\right|\) \(\theta=\tan ^{-1}\left(\frac{24}{7 \sqrt{5}}\right)\)
JEE Main-2022-26.07.2022
Application of Derivatives
85788
The acute angle between the tangents drawn at the point of intersection (other than the origin) of the curves \(x^{2}=4 y\) and \(y^{2}=4 x\) is
1 \(\tan ^{-1}\left(\frac{1}{2}\right)\)
2 \(\sin ^{-1}\left(\frac{3}{5}\right)\)
3 \(\cos ^{-1}\left(\frac{1}{3}\right)\)
4 \(\tan ^{-1}\left(\frac{2}{3}\right)\)
Explanation:
(B) : Given that, \(y^{2}=4 x\) Differentiating w.r.t. \(\mathrm{x}\), we get - \(2 y \frac{d y}{d x}=4\) \(\frac{d y}{d x}=\frac{2}{y} \tag{i}\) And, \(\quad x^{2}=4 y\) Differentiating w.r.t. \(\mathrm{x}\), we get - \(2 x=4 \frac{d y}{d x}\) \(\frac{d y}{d x}=\frac{x}{2} \tag{ii}\) Then, the slope of tangent at \((0,0)\) of equation (i) is parallel to y axis. And, the slope of tangent at \((0,0)\) of equation (ii) parallel to \(\mathrm{x}\) - axis. But, given the point of intersection (other than the origin) Then, let the slope of tangent at \((1,1)\) of equation (i)- \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}_{1}=\frac{2}{1}=2\) And, the slope of tangent at \((1,1)\) of equation (ii) is, \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{m}_{2}=\frac{1}{2}\) Then, \(\theta\) be the angle between the tangent \(\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \Rightarrow \tan \theta=\left|\frac{2-\frac{1}{2}}{1+2 \times \frac{1}{2}}\right|\) \(\tan \theta=\left|\frac{\frac{3}{2}}{1+1}\right| \Rightarrow \tan =\frac{3}{4}\) Then, \(\quad \sin \theta=\frac{3}{5}\) So, \(\theta=\sin ^{-1}\left(\frac{3}{5}\right)\)
AP EAMCET-2018-22.04.2018
Application of Derivatives
85789
Two sides of a triangle are given. If the area of the triangle is maximum then the angle between the given sides is
1 \(45^{\circ}\)
2 \(30^{\circ}\)
3 \(60^{\circ}\)
4 \(90^{\circ}\)
Explanation:
(D) : Let, \(x\) and \(y\) be the length of given sides and \(\theta\) be angle between them. Let, \(A\) be the area of the triangle. Then, \(\quad A=\frac{1}{2} x y \sin \theta \Rightarrow A(\theta)=\frac{1}{2} x y \sin \theta\) \(A^{\prime}(\theta)=\frac{1}{2} x y \cos \theta \Rightarrow A^{\prime \prime}(\theta)=-\frac{1}{2} x y \sin \theta\) For maxima \(/\) minima \(\mathrm{A}^{\prime}(\theta)=0\) \(\frac{1}{2}(x y \cos \theta)=0\) \(\cos \theta=0\) \(\theta=\frac{\pi}{2}\) \(A^{\prime \prime}\left(\frac{\pi}{2}\right)=-\frac{1}{2} x y \sin \frac{\pi}{2}=-\frac{1}{2} x y\lt 0\) So, area is maximum at \(\theta=\frac{\pi}{2}\) or \(90^{\circ}\)
AP EAMCET-2022-06.07.2022
Application of Derivatives
85790
The angle between the tangents drawn at \((0,0)\) to the curves \(y^{3}-x^{2} y+5 y-2 x=0\) and \(x^{4}-x^{3} y^{2}+5 x+2 y=0\) is
1 \(\pi / 6\)
2 \(\pi / 4\)
3 \(\pi / 3\)
4 \(\pi / 2\)
Explanation:
(D) : Given that, \(y^{3}-x^{2} y+5 y-2 x=0\) Slope of tangent to curve at \((10,0)\) \(\mathrm{m}_{1}=\frac{2}{5}\) For second curve is \(x^{4}-x^{3} y^{2}+5 x+2 y=0\) Slope of tangent to curve at \((0,0)\) \(\mathrm{m}_{2}=-\frac{5}{2}\) Therefore, \(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=\frac{2}{5} \times \frac{-5}{2}=-1\) So, angle between tangent to the curves is \(\frac{\pi}{2}\).
