Find the equation of a circle of radius
Question: Find the equation of a circle of radius 5 which is touching another circle x2+ y2 2x 4y 20 = 0 at (5, 5). Solution: Given x2 2x + y2 4y 20 = 0 x2 2x + 1 +y2 4y + 4 20 5 = 0 (x 1)2+ (y 2)2= 25 (x 1)2+ (y 2)2= 52 Since, the equation of a circle having centre (h, k), having radius as r units, is (x h)2+ (y k)2= r2 Centre = (1, 2) Point of Intersection = (5, 5) It intersects the line into 1: 1, as the radius of both the circles is 5 units. Using Ratio Formula, $\frac{\mathrm{m}_{1} \mathrm...
Read More →The length of the major axis of an ellipse is 20 units, and its foci
Question: The length of the major axis of an ellipse is 20 units, and its foci are $(\pm 5 \sqrt{3}, 0)$ Find the equation of the ellipse. Solution: Let the equation of the required ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ Given: Length of Major Axis = 20units ...(i) We know that, Length of Major Axis $=2 \mathrm{a}$...(ii) From eq. (i) and (ii), we get $2 a=20$ $\Rightarrow a=10$ It is also given that, Coordinates of foci $=(\pm 5 \sqrt{3}, 0)$...(iii) We know that, Coordinates of...
Read More →Prove that the semi-vertical angle
Question: Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface is cot $^{-1}(\sqrt{2})$. Solution: Let: Radius of the base $=r$, Height $=h$, Slant height $=1$, Volume $=V$, Curved surface area $=C$ As, Volume, $V=\frac{1}{3} \pi r^{2} h$ $\Rightarrow h=\frac{3 V}{\pi r^{2}}$ Also, the slant height, $l=\sqrt{h^{2}+r^{2}}$ $=\sqrt{\left(\frac{3 V}{\pi r^{2}}\right)^{2}+r^{2}}$ $=\sqrt{\frac{9 V^{2}}{\pi^{2} r^{4}}+r^{2}}$ $=\sqrt{\frac{9 V^{2}+\pi...
Read More →Find the equation of a circle whose centre is (3, –1)
Question: Find the equation of a circle whose centre is (3, 1) and which cuts off a chord of length 6 units on the line 2x 5y + 18 = 0. Solution: Given equation of the chord is, $2 x-5 y+18=0$ $5 y=2 x+18$ $y=\frac{2}{5} x+\frac{18}{5}$ As we have, $y=m x+C$ Where, $\mathrm{m}$ is the slope of the line, $m=\frac{2}{5}$ Slope of the line perpendicular to the chord, $m^{\prime}=-\frac{5}{2}$ As the product of slope of perpendicular lines $=-1$, $y-y_{1}=m^{\prime}\left(x-x_{1}\right)$ $y-(-1)=-\fr...
Read More →Find the equation of the ellipse the ends of whose major and minor axes
Question: Find the equation of the ellipse the ends of whose major and minor axes are (4, 0) and (0, 3) respectively. Solution: Given: Ends of Major Axis $=(\pm 4,0)$ and Ends of Minor Axis $=(0, \pm 3)$ Here, we can see that the major axis is along the $x-$ axis. $\therefore$ The Equation of Ellipse is of the form, $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ (i) where, a is the semi major axis and b is the semi minor axis. Accordingly, $a=4$ and $b=3$ Substituting the value of $a$ and $b$ in eq...
Read More →Show that the cone of the greatest volume
Question: Show that the cone of the greatest volume which can be inscribed in a given spher has an altitude equal to $2 / 3$ of the diameter of the sphere. Solution: Let $h, r$ and $R$ be the height, radius of base of the cone and radius of the sphere, respectively. Then, $h=R+\sqrt{R^{2}-r^{2}}$ $\Rightarrow(h-R)^{2}=R^{2}-r^{2}$ $\Rightarrow h^{2}+R^{2}-2 h r=R^{2}-r^{2}$ $\Rightarrow r^{2}=2 h R-h^{2}$ ....(1) Volume of cone $=\frac{1}{3} \pi r^{2} h$ $\Rightarrow V=\frac{1}{3} \pi h\left(2 h...
Read More →Find the equation of the circle which passes
Question: Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y 4x + 3 = 0. Solution: Since, the equation of a circle having centre (h, k), having radius as r units, is (x h)2+ (y k)2= r2..1 Substituting (2, 3) (4, 5) in the above equation, we get (2 h)2+ (3 k)2= r2 4 4h + h2+ 9 + k2 6k = r2 h2 4h + k2 6k + 13 = r2..2 (4 h)2+ (5 k)2= r2 16 8h + h2+ 25 + k2 10k = r2 h2 8h + k2 10k + 41 = r2..3 Equating both the equations 2 3, ...
Read More →If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters
Question: If the lines 2x 3y = 5 and 3x 4y = 7 are the diameters of a circle of area 154square units, then obtain the equation of the circle. Solution: Since, diameters of a circle intersect at the centre of a circle, 2x 3y = 5 1 3x 4y = 7 ..2 Solving the above equations, Multiplying equation 1 by 3 we get 6x 9y = 15 Multiplying equation 2 by 2 we get 6x 8y = 14 y = 1 y = -1 Putting y = -1, in equation 1, we get 2x 3(-1) = 5 2x + 3 = 5 2x = 2 x = 1 Coordinates of centre = (1,-1) Given area = 154...
Read More →Find the equation of the ellipse whose vertices are the (0, ±4) and foci
Question: Find the equation of the ellipse whose vertices are the (0, 4) and foci at $(0, \pm \sqrt{7})$ Solution: Given: Vertices $=(0, \pm 4) \ldots$ (i) The vertices are of the form $=(0, \pm a) \ldots$ (ii) Hence, the major axis is along $y$-axis $\therefore$ From eq. (i) and (ii), we get $a=4$ $\Rightarrow a^{2}=16$ and We know that, if the major axis is along y axis then the equation of Ellipse is of the form of $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$ Also, given coordinate of foci $=(...
Read More →Find the equation of the hyperbola
Question: Find the equation of the hyperbola with eccentricity 3/2 and foci at ( 2, 0). Solution: Given $e=\frac{3}{2}$ We have foci $=(\pm \mathrm{a} \mathrm{e}, 0)=(\pm 2,0)$ Therefore the hyperbola lies on $x$-axis, Equation is $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ Given $\mathrm{a} \mathrm{e}=2$ $a \times \frac{3}{2}=2$ $a=\frac{4}{3}$ $\because b^{2}=a^{2}\left(e^{2}-1\right)$ $b^{2}=\left(\frac{4}{3}\right)^{2}\left(\left(\frac{3}{2}\right)^{2}-1\...
Read More →Find the equation of the ellipse whose vertices are at
Question: Find the equation of the ellipse whose vertices are at (6, 0) and foci at (4, 0). Solution: Given: Vertices $=(\pm 6,0) \ldots$ (i) The vertices are of the form $=(\pm a, 0) \ldots$ (ii) Hence, the major axis is along $x$-axis $\therefore$ From eq. (i) and (ii), we get $a=6$ $\Rightarrow a^{2}=36$ and We know that, if the major axis is along x axis then the equation of Ellipse is of the form of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ Also, given coordinate of foci $=(\pm 4,0) \ldot...
Read More →If the distance between the foci
Question: If the distance between the foci of a hyperbola is 16 and its eccentricity is 2, then obtain the equation of the hyperbola. Solution: We know that equation of Hyperbola $=\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ Also we have foci $=(\pm a e, 0)$ Given distance between foci is $2 \mathrm{ae}=16$ $e=\sqrt{2}$ $2 \times a \times \sqrt{2}=16$ $a=\frac{16}{2 \times \sqrt{2}}=\frac{8}{\sqrt{2}}=4 \sqrt{2}$ $\because b^{2}=a^{2}\left(e^{2}-1\right)$ $\mathrm{b}^{2}=(4 \sqrt{2})^{2}\left((\s...
Read More →Prove that a conical tent of given
Question: Prove that a conical tent of given capacity will require the least amount of canavas when the height is $\sqrt{2}$ times the radius of the base. Solution: Let the surface area of conical tent be $S=\pi \mathrm{r} \sqrt{\mathrm{r}^{2}+\mathrm{h}^{2}}$ Let the volume of the conical tent $\mathrm{V}=\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h}$ $\Rightarrow \mathrm{h}=\frac{3 \mathrm{~V}}{\pi \mathrm{r}^{2}}$ $\therefore S=\pi r \sqrt{r^{2}+\left(\frac{3 V}{\pi r^{2}}\right)^{2}}$ $\Rightar...
Read More →If the line y = mx + 1 is tangent
Question: If the line y = mx + 1 is tangent to the parabola y2 = 4x then find the value of m. Solution: Given equations are, y = mx + 1 y2= 4x By solving given equations we get (mx + 1)2= 4x Expanding the above equation we get m2x2+ 2mx + 1 = 4x On rearranging we get m2x2+ 2mx 4x + 1 = 0 m+x2+ x (2m 4) + 1 = 0 As the line touches the parabola, above equation must have equal roots, Discriminant (D) = 0 (2m 4)2 4 (m2) (1) = 0 4m2 16m + 16 4m2= 0 -16 m + 16 = 0 m + 1 = 0 m = 1 Hence, the required v...
Read More →If the points (0, 4) and (0, 2) are respectively
Question: If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola. Solution: Given Vertex $=(0,4)$ Focus $=(0,2)$ So, the directrix of the parabola is $y=6$, Since, distance of $(x, y)$ from $(0,2)$ and perpendicular distance from $(x, y)$ to directrix are always equal. Using Distance Formula \ Perpendicular Distance Formula, Perpendicular Distance (Between a point and line) is $=\frac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}...
Read More →Find the value
Question: Find the (i) lengths of major axes, (ii) coordinates of the vertices, (iii) coordinates of the foci, (iv) eccentricity, and (v) length of the latus rectum of each of the following ellipses. $25 x^{2}+4 y^{2}=100$ Solution: Given: $25 x^{2}+4 y^{2}=100$ Divide by 100 to both the sides, we get $\frac{25}{100} x^{2}+\frac{4}{100} y^{2}=1$ $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$ (i) Since, $425$ So, above equation is of the form, $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$ (ii) Comparing eq. ...
Read More →Find the length of the line-segment joining
Question: Find the length of the line-segment joining the vertex of the parabola y2 = 4axand a point on the parabola where the line-segment makes an angle q to the x-axis. Solution: We know that equation of an ellipse is $y^{2}=4 a x$, Also we have Length of latus rectum $=4 a$ Let the point on parabola be (i, j) From the figure, slope of $O P=\tan \theta=\frac{j}{i}$ $\mathrm{j}=i \tan \theta$ Squaring the above equation on both sides we get $j^{2}=i^{2} \tan ^{2} \theta$ $j^{2}=4 a i$ $O P=\sq...
Read More →Find the coordinates of a point on
Question: Find the coordinates of a point on the parabola y2 = 8x whose focal distance is 4. Solution: We know that equation of an ellipse is y2= 4ax, Also we have length of latus rectum = 4a Now by comparing the above two equations, 4a = 8 Therefore a = 2 y2= 8 2 = 16 y = 4 and x = 2 Hence the co ordinates are (2, 4) and (2, -4)....
Read More →Find the distance between the directrices
Question: Find the distance between the directrices of the ellipse x2/36 + y2/20 = 1 Solution: We know that equation of an ellipse $=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ Also we have Length of latus rectum $=\frac{2 b^{2}}{a}$ Length of minor axis $=2 \mathrm{~b}$ $\mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right) \ldots \ldots \ldots \ldots 1$ Now by substituting we get $\frac{x^{2}}{(\sqrt{36})^{2}}+\frac{y^{2}}{(\sqrt{20})^{2}}=1$ $\frac{x^{2}}{(6)^{2}}+\frac{y^{2}}{(2 \sqrt{5}...
Read More →Find the equation of ellipse whose eccentricity is 2/3 ,
Question: Find the equation of ellipse whose eccentricity is 2/3 , latus rectum is 5 and the centre is (0, 0). Solution: We know that equation of an ellipse $=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ Also we have, Length of latus rectum $=\frac{2 b^{2}}{a}$ Length of minor $a x i s=2 b$ Given $e=\frac{2}{3}$ $\mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right) \ldots \ldots \ldots 1$ Length of Latus Rectum $=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=5$ $\mathrm{b}^{2}=2.5 \mathrm{a} \ldots \...
Read More →A rectangle is inscribed in a semi-circle
Question: A rectangle is inscribed in a semi-circle of radius r with one of its sides on diameter of semi-circle. Find the dimension of the rectangle so that its area is maximum. Find also the area. Solution: Let the dimensions of the rectangle be $x$ and $y$. Then, $\frac{x^{2}}{4}+y^{2}=r^{2}$ $\Rightarrow x^{2}+4 y^{2}=4 r^{2}$ $\Rightarrow x^{2}=4\left(r^{2}-y^{2}\right)$ .....(1) Area of rectangle $=x y$ $\Rightarrow A=x y$ Squaring both sides, we get $\Rightarrow A^{2}=x^{2} y^{2}$ $\Right...
Read More →If the eccentricity of an ellipse is 5/8
Question: If the eccentricity of an ellipse is 5/8 and the distance between its foci is 10, then find latus rectum of the ellipse. Solution: We know that equation of an ellipse $=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ Also we have Length of latus rectum $=\frac{2 b^{2}}{a}$ Length of minor axis $=2 \mathrm{~b}$ $e=\frac{5}{8}$ We know that foci $=(\pm a e, 0)$ Given that distance between foci $=10$ $2 \mathrm{ae}=10$ $2 \times a \times \frac{5}{8}=10$ $a=8$ $b^{2}=a^{2}\left(1-e^{2}\right)$ ...
Read More →Show that the height of the cylinder of maximum volume
Question: Show that the height of the cylinder of maximum volume that can be inscribed a sphere of radius $R$ is $\frac{2 R}{\sqrt{3}}$. Solution: Let the height and radius of the base of the cylinder be $h$ and $r$, respectively. Then, $\frac{h^{2}}{4}+r^{2}=R^{2}$ $\Rightarrow h=2 \sqrt{R^{2}-r^{2}}$ ......(1) Volume of cylinder, $V=\pi r^{2} h$ Squaring both sides, we get $\Rightarrow V^{2}=\pi^{2} r^{4} h^{2}$ $\Rightarrow V^{2}=4 \pi^{2} r^{4}\left(R^{2}-r^{2}\right)$ [From eq. (1)] Now, $Z...
Read More →Given the ellipse with equation
Question: Given the ellipse with equation 9x2+ 25y2= 225, find the eccentricity and foci. Solution: We know that equation of an ellipse is $=\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ Also we have, Length of latus rectum $=\frac{2 b^{2}}{a}$ Length of minor axis $=2 b$ $9 x^{2}+25 y^{2}=225$ Dividing the above equation by 225 , $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ The above equation can be written as $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$ From the above ...
Read More →A large window has the shape of a rectangle surmounted by an equilateral triangle.
Question: A large window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 metres find the dimensions of the rectangle will produce the largest area of the window. Solution: Let the dimensions of the rectangle be $x$ and $y$. Perimeter of the window $=x+y+x+x+y=12$ $\Rightarrow 3 x+2 y=12$ $\Rightarrow y=\frac{12-3 x}{2}$ ......(1) Area of the window $=x y+\frac{\sqrt{3}}{4} \mathrm{x}^{2}$ $\Rightarrow A=x\left(\frac{12-3 x}{2}\right)+\frac...
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