Home
Class 12
MATHS
There is a point (p,q) on the graph of f...

There is a point (p,q) on the graph of `f(x)=x^2` and a point `(r , s)` on the graph of `g(x)=(-8)/x ,w h e r ep >0a n dr > 0.` If the line through `(p , q)a n d(r , s)` is also tangent to both the curves at these points, respectively, then the value of `P+r` is_________.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the value of \( p + r \) given the conditions about the points on the curves \( f(x) = x^2 \) and \( g(x) = -\frac{8}{x} \). ### Step 1: Identify the points on the curves The point \( (p, q) \) lies on the graph of \( f(x) = x^2 \). Therefore, we can express \( q \) in terms of \( p \): \[ q = f(p) = p^2 \] The point \( (r, s) \) lies on the graph of \( g(x) = -\frac{8}{x} \). Thus, we can express \( s \) in terms of \( r \): \[ s = g(r) = -\frac{8}{r} \] ### Step 2: Find the slopes of the tangents Next, we find the derivatives of both functions to determine the slopes of the tangents at points \( (p, q) \) and \( (r, s) \). The derivative of \( f(x) \) is: \[ f'(x) = 2x \] At \( x = p \): \[ f'(p) = 2p \] The derivative of \( g(x) \) is: \[ g'(x) = \frac{8}{x^2} \] At \( x = r \): \[ g'(r) = \frac{8}{r^2} \] ### Step 3: Set the slopes equal Since the line through the points \( (p, q) \) and \( (r, s) \) is tangent to both curves, the slopes must be equal: \[ 2p = \frac{8}{r^2} \] ### Step 4: Solve for \( p \) in terms of \( r \) Rearranging the equation gives: \[ p = \frac{4}{r^2} \] ### Step 5: Find the slope of the line through the two points The slope of the line through the points \( (p, q) \) and \( (r, s) \) can be expressed as: \[ \text{slope} = \frac{s - q}{r - p} \] Substituting for \( s \) and \( q \): \[ \text{slope} = \frac{-\frac{8}{r} - p^2}{r - p} \] ### Step 6: Set the slopes equal Setting this equal to \( 2p \): \[ \frac{-\frac{8}{r} - p^2}{r - p} = 2p \] ### Step 7: Substitute \( p \) into the equation Substituting \( p = \frac{4}{r^2} \) into the equation: \[ \frac{-\frac{8}{r} - \left(\frac{4}{r^2}\right)^2}{r - \frac{4}{r^2}} = 2\left(\frac{4}{r^2}\right) \] ### Step 8: Simplify and solve for \( r \) After simplifying, we can find the value of \( r \). 1. The left-hand side simplifies to: \[ -\frac{8}{r} - \frac{16}{r^4} \] and the denominator becomes: \[ r - \frac{4}{r^2} = \frac{r^3 - 4}{r^2} \] 2. Setting the equation and solving gives us \( r = 1 \). ### Step 9: Find \( p \) Substituting \( r = 1 \) back into \( p = \frac{4}{r^2} \): \[ p = \frac{4}{1^2} = 4 \] ### Step 10: Calculate \( p + r \) Finally, we calculate: \[ p + r = 4 + 1 = 5 \] ### Final Answer The value of \( p + r \) is \( \boxed{5} \).
Promotional Banner

Topper's Solved these Questions

  • APPLICATION OF DERIVATIVES

    CENGAGE ENGLISH|Exercise JEE PREVIOUS YEAR|10 Videos
  • APPLICATION OF DERIVATIVES

    CENGAGE ENGLISH|Exercise ILLUSTRATION|62 Videos
  • APPLICATION OF DERIVATIVES

    CENGAGE ENGLISH|Exercise LINKED COMPREHENSION TYPE|8 Videos
  • 3D COORDINATION SYSTEM

    CENGAGE ENGLISH|Exercise DPP 3.1|11 Videos
  • APPLICATION OF INTEGRALS

    CENGAGE ENGLISH|Exercise All Questions|142 Videos

Similar Questions

Explore conceptually related problems

Write a unit vector in the direction of vec P Q ,\ w h e r e\ P\ a n d\ Q are the points (1, 3, 0) and (4, 5, 6) respectively.

The graph of function f contains the point P(1, 2) and Q(s, r) . The equation of the secant line through P and Q is y=((s^2+2s-3)/(s-1))x-1-s . The value of f'(1) , is

The points (p,q +r) , (q, r+p) and (r, q+p) are "_______" .

In Figure, T is a point on side Q R of P Q R and S is a point such that R T=S Tdot Prove That : P Q+P R > Q S

If points Q and R reflections of point P(-3,\ 4) in X and Y axes respectively, what is Q R ?

A point R with x-coordinates 4 lies on the line segment joining the points P(2,-3,4)a n d\ Q(8,0, 10)dot Find the coordinates of the point Rdot

In Figure, P Q > P R. Q S\ a n d\ R S are the bisectors of /_ Q and /_R respectively. Prove that S Q > S R .

Tangent is drawn at any point (p ,q) on the parabola y^2=4a x .Tangents are drawn from any point on this tangant to the circle x^2+y^2=a^2 , such that the chords of contact pass through a fixed point (r , s) . Then p ,q ,r and s can hold the relation (A) r^2q=4p^2s (B) r q^2=4p s^2 (C) r q^2=-4p s^2 (D) r^2q=-4p^2s

Find the coordinates of points P ,\ Q ,\ R\ a n d\ S in Figure.

If p,q are roots of the quadratic equation x^(2)-10rx -11s =0 and r,s are roots of x^(2)-10px -11q=0 then find the value of p+q +r+s.

CENGAGE ENGLISH-APPLICATION OF DERIVATIVES-NUMERICAL VALUE TYPE
  1. There is a point (p,q) on the graph of f(x)=x^2 and a point (r , s) on...

    Text Solution

    |

  2. A curve is defined parametrically be equations x=t^2a n dy=t^3 . A var...

    Text Solution

    |

  3. If d is the minimum distance between the curves f(x)=e^x a n dg(x)=(lo...

    Text Solution

    |

  4. Let f(x0 be a non-constant thrice differentiable function defined on (...

    Text Solution

    |

  5. At the point P(a , a^n) on the graph of y=x^n ,(n in N), in the first...

    Text Solution

    |

  6. A curve is given by the equations x=sec^2theta,y=cotthetadot If the ta...

    Text Solution

    |

  7. Water is dropped at the rate of 2 m^3/s into a cone of semi-vertical a...

    Text Solution

    |

  8. If the slope of line through the origin which is tangent to the curve ...

    Text Solution

    |

  9. Let y=f(x) be drawn with f(0) =2 and for each real number a the line t...

    Text Solution

    |

  10. Suppose a , b , c are such that the curve y=a x^2+b x+c is tangent to ...

    Text Solution

    |

  11. Let C be a curve defined by y=e^(a+b x^2)dot The curve C passes throug...

    Text Solution

    |

  12. If the curve C in the x y plane has the equation x^2+x y+y^2=1, then t...

    Text Solution

    |

  13. If a , b are two real numbers with a<b then a real number c can be fo...

    Text Solution

    |

  14. Let f:[1,3]to[0,oo) be continuous and differentiabl function. If (f(3)...

    Text Solution

    |

  15. The x intercept of the tangent to a curve f(x,y) = 0 is equal to the o...

    Text Solution

    |

  16. if f(x) is differentiable function such that f(1) = sin 1, f (2)= sin ...

    Text Solution

    |

  17. Let f(x)=x(x^(2)+mx+n)+2," for all" x neR and m, n in R. If Rolle's t...

    Text Solution

    |

  18. If length of the perpendicular from the origin upon the tangent drawn ...

    Text Solution

    |

  19. If f (x) ={{:(xlog(e)x",",x gt0),(0",",x=0):} ,thenconclusion of LMVT ...

    Text Solution

    |