Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Straight Lines and Pair of Straight Lines question

2009 · Shift 0 · Q41
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Straight Lines and Pair of Straight Lines
  5. /2009 · Shift 0 · Q41

Straight Lines and Pair of Straight Lines question

2009 · Shift 0 · Q41

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
The shortest distance between the line y−x=1y - x = 1y−x=1 and the curve x=y2x = {y^2}x=y2 is :
  1. A
    238{{2\sqrt 3 } \over 8}823​​
  2. B
    325{{3\sqrt 2 } \over 5}532​​
  3. C
    34{{\sqrt 3 } \over 4}43​​
  4. D
    328{{3\sqrt 2 } \over 8}832​​
View written solutionFree

Correct answer: D

  1. Rewrite the line in a convenient form

The line is y−x=1  ⟹  y=x+1.y-x=1 \implies y=x+1.y−x=1⟹y=x+1. In standard form: x−y+1=0.x-y+1=0.x−y+1=0.

For any point (x,y)(x,y)(x,y), its perpendicular distance from this line is d=∣x−y+1∣12+(−1)2=∣x−y+1∣2.d=\frac{|x-y+1|}{\sqrt{1^2+(-1)^2}}=\frac{|x-y+1|}{\sqrt2}.d=12+(−1)2​∣x−y+1∣​=2​∣x−y+1∣​.

  1. Take a general point on the curve

The curve is x=y2.x=y^2.x=y2. So a general point on the curve is (y2,y).(y^2,y).(y2,y).

Its distance from the line is d(y)=∣y2−y+1∣2.d(y)=\frac{|y^2-y+1|}{\sqrt2}.d(y)=2​∣y2−y+1∣​.

We need the shortest distance, so minimize ∣y2−y+1∣.|y^2-y+1|.∣y2−y+1∣.

  1. Analyze the expression inside modulus

Consider f(y)=y2−y+1.f(y)=y^2-y+1.f(y)=y2−y+1. Complete the square: f(y)=(y−12)2+34.f(y)=\left(y-\frac12\right)^2+\frac34.f(y)=(y−21​)2+43​.

Hence, f(y)>0for all y,f(y)>0 \quad \text{for all } y,f(y)>0for all y, so the modulus is unnecessary: ∣y2−y+1∣=y2−y+1.|y^2-y+1|=y^2-y+1.∣y2−y+1∣=y2−y+1.

Thus we minimize d(y)=y2−y+12.d(y)=\frac{y^2-y+1}{\sqrt2}.d(y)=2​y2−y+1​.

  1. Find the minimum

Since y2−y+1=(y−12)2+34,y^2-y+1=\left(y-\frac12\right)^2+\frac34,y2−y+1=(y−21​)2+43​, its minimum value is 34,\frac34,43​, achieved at y=12.y=\frac12.y=21​.

Therefore, dmin⁡=342=342.d_{\min}=\frac{\frac34}{\sqrt2}=\frac{3}{4\sqrt2}.dmin​=2​43​​=42​3​. Rationalizing, dmin⁡=328.d_{\min}=\frac{3\sqrt2}{8}.dmin​=832​​.

  1. Match with the options

328\frac{3\sqrt2}{8}832​​ corresponds to Option D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So the stored answer is correct.

PreviousNext

More from Straight Lines and Pair of Straight Lines

  • The perpendicular bisector of the line segment joining P(1, 4) and Q(k, 3) has y-intercept -4. Then a possible value of k is :2008 · MCQ
  • Let A (h,k), B (1,1) and C (2,1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is 1 square unit, then the set of values which ′k′ can take is…2007 · MCQ
  • A straight line through the point A(3,4) is such that its intercept between the axes is bisected at A. Its equation is :2006 · MCQ
  • If (a,a2) falls inside the angle made by the lines y=2x​,x>0 and y=3x,x>0, then a belong to :2006 · MCQ
  • The line parallel to the x- axis and passing through the intersection of the lines ax+2by+3b=0 and bx−2ay−3a=0, where (a,b)e(0,0) is :2005 · MCQ
  • If non zero numbers a,b,c are in H.P., then the straight line ax​+by​+c1​=0 always passes through a fixed point. That point is :2005 · MCQ
  • If a vertex of a triangle is (1,1) and the mid points of two sides through this vertex are (−1,2) and (3,2) then the centroid of the triangle is :2005 · MCQ
  • The equation of the straight line passing through the point (4,3) and making intercepts on the co-ordinate axes whose sum is −1 is :2004 · MCQ